i.^/ 


IN  MEMORIAM 
FLORIAN  CAJORI 


\^f^JL^^^^ 


FIRST  YEAR  MATHEMATICS 


BY 

GEORGE  W.   EVANS 

HEADMASTER    OF    THE    CHARLESTOWN    HIGH    SCHOOL,    BOSTON 

AND 

JOHN  A.   MARSH 

MASTER    IN    MATHEMATICS    AT    THE    ENGLISH    HIGH    SCHOOL, 

BOSTON 


CHARLES   E.   MERRILL   COMPANY 
NEW  YORK  AND   CHICAGO 


COPTKIGHT,    1916, 

By   CHARLES  E.    MERRILL  CO. 


El 


PREFACE 

This  book  is  the  result  of  twenty  years  of  patient  experi- 
ment in  actual  teaching.  It  is  intended  to  be  completed  in 
the  first  year  of  the  high  school.  It  presents  algebraic 
equations  primarily  as  a  device  for  the  solution  of  problems 
stated  in  words,  and  gives  a  complete  treatment  of  numerical 
equations  such  as  are  usually  included  in  high-school  algebra 
—  one-letter  and  two-letter  equations,  integral  and  frac- 
tional, including  one-letter  quadratics  and  the  linear-quad- 
ratic pair.  So  much  of  algebraic  manipulation  is  included 
as  is  necessary  for  th^e  treatment  of  these  equations. 

The  arithmetic  in  the  book  is  presented  from  a  new  point 
of  view  —  that  of  approximate  computation  —  and  is  utilized 
in  the  evaluation  of  formulas  and  in  the  solution  of  equations 
throughout  the  succeeding  pages. 

Geometrical  facts  are  introduced  as  the  basis  of  many 
algebraic  and  arithmetic  problems,  and  wherever  they  are 
not  intuitively  accepted  by  the  pupils  they  are  accompanied 
by  adequate  logical  demonstration.  Proofs,  and  parts  of 
proofs,  are  avoided  when  they  seem  to  the  pupils  of  an 
unnecessary  and  hair-splitting  kind. 

AH  problems  are  carefully  graded,  for  it  is  by  means  of 
problems  that  each  successive  algebraic  difficulty  is  in- 
troduced. 

A  great  deal  of  pains  has  been  taken  to  present  new  topics 
clearly  and  concretely,  often  dividing  them  into  sub-topics 
each  of  which  is  separately  illustrated  and  apphed  to  prac- 
tice. Definitions  are  generally  prepared  for  by  such  advance 
work  as  will  cause  the  student  to  feel  the  need  of  them;  and 
where  no  need  exists,  they  are  omitted.  To  introduce  defi- 
nitions of  concepts  that  are  already  famifiar  to  the  student, 

3 


t#^^^^  #^^ 


4  PREFACE 

such  as  angle  and  right  angle,  would  increase  the  formalism 
of  the  exposition,  and  be  on  the  whole  repellant. 

The  book  has  been  used  in  manuscript,  and  substantially 
in  its  present  form,  for  four  years  in  one  of  the  largest  high 
schools  in  Boston.  It  has  been  found  workable.  It  gives 
the  pupil  who  has  but  one  year  for  mathematics  a  substantial 
knowledge  of  the  meaning  and  purpose  of  algebra,  and  more 
than  a  hint  of  what  geometry  is  about.  Neither  of  these 
things  can  be  said  of  the  conventional  first  year  mathematics. 
Pupils  who  go  on  with  mathematics,  after  the  course  covered 
by  this  textbook,  enter  upon  their  second  year  with  habits  of 
self-reliance  and  of  self-criticism,  and  complete  the  usual 
syllabuses  of  college  preparation  without  duplication  of  work 
or  loss  of  time.  Teachers  who  have  begun  the  text  with 
some  distrust  have  finished  it  with  enthusiasm. 

Other  features  of  the  book  are: 

1.  The  method  of  solution  of  equations:  the  logical 
explanation  of  each  step  is  noted  in  writing,  with  a  clear  and 
systematic  symbolization,  by  the  pupil  himself.  Technical 
terms  of  riianipulation,  such  as  "transposing"  and  "clearing 
of  fractions,"  are  replaced  by  briefer  statements  of  the  actual 
thing  done  to  each  of  the  equal  numbers. 

2.  Approximate  computation  is  made  not  only  a  time- 
saver  but  a  training  in  common  sense. 

3.  The  graphical  method  is  used  not  merely  as  a  prepara- 
tion for  examination  questions  but  as  a  means  of  exhibiting 
the  functional  relation  of  the  two  letters  of  the  equations 
used,  and  as  a  means  of  clarifjdng  the  study  of  elimination; 
it  is  made  an  integral  part  of  the  treatment  of  that  subject. 

4.  The  solution  of  quadratics  is  treated  by  the  general 
methods  of  algebra,  instead  of  using  a  method  pecuUar  to 
high-school  work.  "Completing  the  square"  is  regarded  as 
a  method  of  factoring  a  quadratic  expression;  and  the  sign 
=b  for  a  square  root  is  deduced  from  the  solution  of  the 
equation  x^  =  n,  instead  of  the  reverse.  Problems  are 
chosen,  first,  to  show  the  answers  distinct  and  of  equal  sig- 


PREFACE  6 

nificance,  then  successively  to  show  them  alike,  apparently 
different  but  really  alike,  and  so  on.  The  algebraic  negative 
appears  here,  with  its  interpretation.  Meaningless  answers 
are  found,  which  are  to  be  discarded;  irrational  answers  are 
exemplified;  imaginary  answers  are  mentioned  as  an  indi- 
cation of  some  incongruity  in  the  statement  of  the  problem. 

5.  Two  methods  of  ehmination  are  given:  "by  combina- 
tion" for  linear  pairs,  and  "by  substitution"  for  linear- 
quadratic  pairs. 

6.  Throughout  the  book,  the  pupil  is  required  to  check  all 
his  work,  and  so  to  rely  upon  himself  for  certitude  of  his 
accuracy. 

7.  The  unusual  arrangement  of  topics  (see  Table  of 
Contents)  and  the  introduction  of  material  from  arithmetic, 
geometry,  and  even  from  trigonometry,  have  been  found  to 
arouse  an  unusual  interest  in  the  pupil  and  to  leave  him  at 
the  end  of  the  year  with  a  feeling  of  achievement  that  pupils 
never  get  from  the  mere  manipulation  of  algebraic  expres- 
sions. 

For  all  the  foregoing  reasons  it  is  believed  that  this  book 
will  disarm  many  of  the  attacks  now  being  made  on  the 
teaching  of  algebra  in  high  schools. 

We  are  greatly  indebted  to  the  wisdom  and  enthusiasm  of 
Mr.  Henry  M.  Wright  and  Mr.  Bertram  C.  Richardson,  who 
have  given  invaluable  aid  in  working  out  the  teaching  sys- 
tem embodied  in  this  book.  Many  other  Boston  teachers, 
notably  Mr.  Walter  F.  Downey  and  Mr.  Peter  F.  Gartland, 
have  generously  cooperated  in  testing  the  various  stages  of 
the  manuscript.  In  offering  the  book  to  a  larger  public, 
therefore,  the  authors  feel  warranted  in  claiming  that  it  is 
representative  of  a  considerable  group  of  scholarly  and  pro- 
gressive teachers. 

GEORGE  W.  EVANS 

JOHN  A.  MARSH 

Boston,  June  1,  1916. 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/firstyearmathemaOOevanrich 


\  CONTENTS 

Chapteb  Page 

I.  The  First  Use  of  Algebra 11 

Equations 13 

Ratio 16 

Measurement  of  Length 17 

Measurement  of  Angles 21 

Valuation  Problems 26 

Difiference  of  the  Unknown  Numbers  Given 30 

Angles  and  Angle  Relations 32 

Multiplications 36 

Shortages 37 

The  Study  of  a  Simple  Equation 41 

Sum  of  the  Unknown  Numbers  Given 44 

Circles 46 

II.  Approximate  Computation 50 

Multiplication 50 

Division 54 

Square  Root 57 

III.  Measurement  Formulas 63 

Rectangles 63 

Rhomboids 66 

Triangles 70 

Trapezoids 75 

The  Number  tt 78 

Implicit  Formulas 80 

IV.  Additional  Types  of  Problems 83 

Fractional  Equations 83 

Subtracting  Fractions  and  Parentheses 88 

The  Construction  of  Equations 91 

The  Problem  of  the  Digits 92 

The  Problem  of  Two  Velocities 94 

The  Problem  of  AbiUty  and  Time 100 

7 


8  CONTENTS 

CHAPTEB  PAGE 

V.  Transformations 103 

Negative  Numbers 103 

Addition 104 

Subtraction 106 

Parentheses  within  Parentheses 108 

Multiplication 109 

The  Distributive  Law 112 

Distributive  Factoring 113 

MultipUcation  of  Polynomials 114 

Division 116 

Long  Division 117 


VI.  Identities  and  Theorems;  Factoring 119 

The  Proof  of  Theorems 119 

Quadratic  Products 121 

Factoring  by  Cross  Multiplication 123 

Completing  the  Square 126 

VII.  Quadratic  Equations 130 

Two  Answers  to  One  Question 130 

Both  Answers  Alike 134 

Answers  Apparently  Different 135 

The  Meaning  of  Negative  Answers 136 

Answers  Suggesting  Related  Problems 138 

Meaningless  Answers 140 

Equations  Involving  Irrational  Numbers 142 

Peculiar  Quadratics 144 

VEIL  Some  Theorems  of  Geometry 147 

Some  Ratios  of  Area 148 

Similar  Figures 155 

Similar  Right  Triangles 159 

Pythagorean  Theorem 162 

Angles  of  a  Polygon 164 

IX.  The  Graphical  Method 167 

The  Algebraic  Scale 167 

The  Algebraic  Diagram ...    170 

The  Locus  of  an  Equation 175 


CONTENTS  9 

CHAPTER  PAGE 

X.   Elimination  by  Combination. 181 

Two-Letter  Problems 182 

Solving  for  Reciprocals 194 

Where  Elimination  Fails 197 

Non-Algebraic  Conditions 198 

XI.  Elimination  by  Substitution ...  200 

Linear-Quadratic  Pairs 200 

Algebraic  Diagrams 206 

The  Equation  of  the  Circle 208 

Other  Quadratic  Loci 210 

The  Standard  Parabola 215 

XII.  Supplementary  Problems  for  Practice  and  Review  218 


FIRST  YEAR  MATHEMATICS 

CHAPTER  ^  I 
THE   FIRST  USE  OF  ALGEBRA 

1.  Algebra  is  a  method  of  abbreviating  the  explanations 
of  arithmetic  problems.  This  was,  historically,  its  first  use. 
It  is  also  used  for  abbreviating  the  statement  of  rules  in 
arithmetic,  and  for  the  discussion  of  numbers  in  every  case 
where  the  numbers  are  described  in  words  instead  of  being 
expressed  in  figures. 

2.  This  chapter  illustrates  the  first  use  of  algebra:  for 
abbreviating  the  explanation  of  problems  in  arithmetic. 
In  the  problems  which  are  here  given  as  illustrations,  the 
answers  are  sometimes  evident  at  a  glance.  The  student  is 
not  expected  to  shorten  the  arithmetical  work,  for  he  can- 
not do  that;  he  is  to  put  down  the  explanation  briefly  and 
systematically.  When  that  is  learned,  he  will  find  that  he 
can  do  complicated  problems  with  greater  ease,  because  he 
can  put  down  in  black  and  white,  as  he  goes  along,  the  suc- 
cessive steps  of  his  reasoning. 

3.  Model  A.  —  A  father  is  6  times  as  old  as  his  son,  and 
their  united  ages  are  42  years.     Find  the  age  of  each. 

Explanation 

(1)  The  father's  age  +  the  son's  age  =  42  years. 

(2)  The  father's  age  =  6  times  the  son's  age. 

(3)  6  times  the  son's  age  +  the  son's  age  =  42  years. 

(4)  7  times  the  son's  age  =  42  years. 

(5)  The  son's  age  =  6  years. 

(6)  The  father's  age  =  36  years. 

11 


12  THE  FIRST  USE  OF  ALGEBRA 

Abbreviated  Explanation 

Let  s  stand  for  the  number  of  years  in  the  son's  age. 

Then  6  X  s  will  stand  for  the  number  of  years  in  the  father's  age. 

©     6xs  +  s  =  42 

@  7  X  s  =  42 

®  s  =    6 

0  6  X  s  =  36 

Ans.    The  father  is  36  years  old;  the  son  is  6  years  old. 

EXERCISE  1 

1.  John  is  twice  as  old  as  Henry,  and  the  sum  of  their 
ages  is  21  years.     Find  the  age  of  each. 

2.  John  and  Will  wish  to  divide  36  cents  so  that  John 
shall  have  twice  as  much  as  Will.  How  much  money  will 
each  boy  have? 

3.  I  paid  $220  for  a  horse  and  saddle,  and  the  horse  cost 
ten  times  as  much  as  the  saddle.  What  was  the  cost  of 
each? 

4.  A  sidewalk  laid  with  a  flagstone  and  a  curbstone  is  6  feet 
wide;  the  flagstone  is  11  times  as  wide  as  the  curbstone. 
What  is  the  width  of  the  flagstone  and  of  the  curb? 

5.  The  electoral  vote  of  the  state  of  New  York  is  9  times 
that  of  the  state  of  South  Dakota.  Both  states  together 
have  50  electoral  votes.  How  many  votes  does  each  state 
have? 

6.  There  are  7  times  as  many  sheep  as  lambs  in  a  pasture, 
and  in  all  there  are  96.     Find  the  number  of  each. 

7.  A  pole  9  feet  long  is  partly  peeled  of  bark;  the  bark 
part  is  twice  as  long  as  the  bare  part.  Find  the  length  of 
each  part. 

8.  It  takes  a  man  four  tunes  as  long  to  run  the  first  three- 
quarters  of  a  measured  mile  as  to  run  the  last  quarter.  He 
goes  the  whole  mile  in  six  minutes.  How  long  does  it  take 
him  to  run  the  last  quarter? 


EQUATIONS  13 

9.  Mercury  weighs  13  times  as  much  as  water;  a  quart 
of  mercury  and  a  quart  of  water  together  weigh  28  pounds. 
What  is  the  weight  of  each? 

Equations 

4.  An  equation  is  a  statement  that  two  numbers  are 
equal. 

5.  The  algebraic  abbreviation  for  any  number,  or  for 
any  combination  of  numbers,  is  called  an  algebraic  ex- 
pression. 

6.  It  takes  two  numbers  to  form  an  equation.  One  or 
both  of  the  numbers  may  be  an  algebraic  expression.  The 
two  numbers  which  are  said  to  be  equal  are  called  the  first 
(left-hand)  member  and  the  second  (right-hand)  member 
of  the  equation. 

7.  The  parts  of  an  algebraic  expression  separated  by  the 
+  and  —  signs  are  called  the  terms  of  the  expression. 

The  statements  numbered  ©,  0,  ®,  0  in  Model  A  (page  12)  are 
equations,  because  each  states  that  two  numbers  are  equal.  In  equation 
0,  the  number  obtained  by  adding  6  X  s  to  s  is  said  to  be  equal  to 
the  number  42. 

There  are  three  terms  in  this  equation:  the  first  is  6  X  s,  and  the 
second  is  s.     These  terms  are  separated  by  the  sign  +. 

Either  of  these  terms  separately,  as  well  as  the  sum  of  both,  is  an 
algebraic  expression.  The  second  member  contains  only  one  term,  a 
number  without  any  letter. 

8.  In  order  to  fix  the  mind  on  the  nature  of  the  processes 
by  which  his  equations  are  obtained,  it  is  well  for  the  student 
to  indicate  beside  each  equation  how  it  arose  from  the  pre- 
ceding equations.  With  very  little  trouble,  he  is  thus  able 
to  complete  his  explanation. 

9.  Model  B.  —  In  a  certain  collection  of  insects,  there 
are  twice  as  many  butterflies  as  moths;  of  both  there  are 
39.    Find  the  number  of  each. 


14  THE  FIRST  USE  OF  ALGEBRA 


Let 

X  —  the  number  of  moths. 

. 

Then 

2X 

X  =  the  number  of  butterflies. 

© 

2XX  +  X  = 

=  39 

® 

SXx  = 

=  39 

©  same  values 

® 

X  = 

=  13 

®  -^3 

® 

2XX-- 

=  26 

Ans 

®  X  2 
.     13  moths; 

26  butterflies. 

In  Model  B,  the  numbers  in  equation  ©  are  obtained 
from  ®  without  any  change  of  value.  The  numbers  in  @ 
are  obtamed  from  the  numbers  in  ©  by  dividing  each  of 
them  by  3,  this  fact  being  indicated  on  the  right  of  ©  by 
the  history  of  the  equation,  ©  -r-  3.  The  numbers  in  © 
are  obtained  from  those  in  ©  by  multiplying  each  of  them 
by  2,  the  history  being,  ©  X  2. 

10.  When  two  number  symbols  are  written  together 
without  any  sign  between  them,  multiplication  is  indicated. 

For  representing  numbers  of  which  values  are  not  stated, 
sometimes  initial  letters  are  used  as  in  Model  A;  more 
often  we  use  a  small  letter  from  the  latter  part  of  the  alpha- 
bet, such  as  X,  y,  or  z. 

These  are  matters  of  accepted  custom  (convention)  among 
writers  and  students  of  algebra. 

11.  The  answer  in  Model  B  is  readily  verified  because  we 
see  at  once  that  the  statements  of  the  problem  are  true  for 
the  numbers  obtained  as  answers.  That  is,  26  =  2  X  13 
and  26  +  13  =  39.     This  verification  is  called  checking. 

Answers  should  always  be  checked  by  referring  to  the 
words  of  the  problem  itself,  not  to  the  first  equation. 

exercise;  2 

1.  One  man  is  twice  as  heavy  as  another,  and  both  weigh 
339  pounds.     Find  the  weight  of  each. 

2.  The  number  of  cows  and  sheep  in  a  certain  farmyard 
is  75,  and  there  are  four  times  as  many  sheep  as  cows.  Find 
the  number  of  each. 


EQUATIONS  15 

3.  The  doctor  has  twice  as  many  books  as  the  minister, 
and  they  both  have  2100.    How  many  books  has  each  man? 

4.  In  a  dog  show  there  are  36  St.  Bernards,  twice  as  many 
long-haired  as  short-haired.  How  many  of  each  kind  are 
there? 

5.  A  locomotive  weighs  6  times  as  much  as  a  car,  and 
both  weigh  14  tons.    Find  the  weight  of  each. 

6.  A  man  is  twice  as  heavy  as  a  boy,  and  both  weigh 
200  pounds.    Find  the  weight  of  each. 

•7.  A's  house  cost  2J  times  as  much  as  B^s  house,  and 
both  cost  $7000.    Find  the  cost  of  each. 

8.  I  paid  $100  for  two  lots  of  tiles;  one  lot  cost  4  times 
as  much  as  the  other.     What  was  the  cost  of  each  lot? 

9.  I  walked  4  times  as  far  in  the  afternoon  as  I  did  in 
the  forenoon;  I  walked  15  miles  altogether.  How  far  did 
I  walk  in  the  forenoon? 

10.  It  costs  4  times  as  much  to  go  from  New  York  to 
Chicago  as  from  Boston  to  New  York;  from  Boston  to 
Chicago,  by  way  of  New  York,  it  costs  $25.  How  much 
does  it  cost  to  go  from  Boston  to  New  York? 

11.  The  president  of  a  stock  company  owns  twice  as  many 
shares  in  it  as  his  brother,  and  both  own  165  shares.  How 
many  shares  does  each  man  own? 

12.  A  man  had  a  big  bill  in  his  pocket,  and  asked  the 
price  of  a  house  and  barn;  he  found  that  the  house  cost  7 
times  the  value  of  the  bill,  and  the  barn  3  times  the  value 
of  the  bill,  both  coming  to  $5000.  Find  the  cost  of  house 
and  barn. 

13.  There  are  three  times  as  many  barrels  in  one  cellar 
as  in  another;  in  the  store  above,  there  are  half  as  many 
barrels  as  in  both  cellars;  altogether  there  are  216  barrels. 
How  many  are  there  in  each  part  of  the  building? 


16  -THE  FIRST  USE  OF  ALGEBRA 

14.  A  fortress  is  garrisoned  by  5200  men;  and  there  j 
9  times  as  many  infantry,  and  3  times  as  many  artillery, 
cavalry.     How  many  are  there  of  each? 

Ratio 

12.  When  one  quantity  can  be  obtained  from  another 
multiplying,  the  multiplier  is  called  their  ratio. 

Thus,  in  Ex.  14  above,  the  ratio  of  infantry  to  cavalry  is 
that  of  cavalry  to  infantry  is  ^.  This  is  onty  another  way 
saying  that  the  number  of  infantry  is  9  times  the  numl 
of  cavalry;  and  that  the  number  of  cavalry  is  ^  the  numl 
of  infantry. 

13.  Model  C.  —  A  line  530  centimeters  long  is  divic 
into  two  portions,  and  the  ratio  of  these  is  4.  Find  the  lenj 
of  each. 

Let  X  =  the  number  of  centimeters  in  one  portion. 

Then    4  x  =  the  number  of  centimeters  in  the  other. 

©    a;  +  4  X  =  530 

0  5  X  =  530        ®  same  values 

(?)  x  =  106        ®  -H  5 

©  4  a;  =  424        0X4 

Ans.    One  portion  is  106  cm.;  the  other  424  cm 

Check:    106  X  4  =    424 

+  106 

530 

EXERCISE  3 

1.  Divide  78  pounds  of  cheese  into  two  portions  whi 
will  have  the  ratio  2. 

2.  Divide  a  farm  of  810  acres  into  two  portions  with 
ratio  8. 

3.  Two  parts  of  a  line  90  feet  long  have  a  ratio  |.  H< 
long  are  the  parts? 


MEASUREMENT  OF  LENGTH  17 

4.  Of  a  200-acre  farm,  part  is  sown  to  one  crop  and  the 
rest  to  another.  The  ratio  of  the  two  parts  is  4.  How 
much  land  is  there  in  each  part? 

5.  The  ratio  of  an  average  man's  weight  to  an  average 
ten-year-old  boy's  weight  is  2.  Such  a  man  and  boy  would 
together  weigh  210  pounds.    What  is  the  weight  of  each? 

6.  Divide  28  into  two  parts  having  a  ratio  6. 

7.  The  ratio  of  the  weight  of  an  average  St.  Bernard  dog 
to  that  of  an  average  Blenheim  spaniel  is  15.  In  the  dog 
show  there  were  10  St.  Bernards  and  5  of  these  spaniels,  and 
altogether  they  weighed  1085  pounds.  What  is  the  average 
weight  of  each  dog? 

8.  The  ratio  of  non-voters  to  voters  in  a  town  is  4.  If 
the  population  is  13,800,  how  many  are  voters? 

9.  The  ratio  of  coin  money  to  bills  in  a  conductor's  pocket 
is  ^;  he  has  $14.40  in  all.  How  much  of  his  money  is  in 
coin? 

10.  In  a  certain  transaction  the  gross  receipts  were  $720. 
The  ratio  of  the  amount  invested  to  the  amount  received  as 
profit  was  8.     How  much  was  invested? 

Measurement  of  Length 

14.  A  straight  line  is  measured  by  marking  off  on  it  a 
certain  standard  straight  Hne  called  a  unit  of  length.  This 
unit  is  in  modern  times  defined  carefully  by  national  govern- 
ments. The  units  in  commonest  use  are  the  foot  and  the 
meter,  but  multiples  and  subdivisions  of  these  units  are  also 
frequently  used. 

When  the  last  measurement  mark  comes  exactly  at  the 
end  of  the  straight  line  we  are  measuring,  the  length-number 
is  a  whole  number,  or  an  integer;  when  it  does  not  come 
exactly  at  the  end  of  the  line,  the  measurement  is  continued 
with  subdivisions  of  the  unit.     The  system  of  subdivision  in 


18  THE  FIRST  USE  OF  ALGEBRA 

ordinary  use  for  the  foot  is  cumbrous,  and  is  not  much  used 
in  careful  work,  such  as  engineering.  Here  decimal  sub- 
division is  abnost  universal. 

15.  The  precision  of  measurement  is  indicated  by  the 
number  of  figures  required  to  express  the  result,  rather  than 
•by  the  number  of  decimal  places.  Thus  a  length  of  88.2 
inches  can  be  expressed,  with  equal  accuracy,  as  7.35  feet,  or 
2.24  meters,  or  even  as  .00139  miles,  depending  upon  the 
unit  used,  and  not  upon  the  care  or  skill  with  which  the 
work  is  done.  These  four  numbers,  representing  the  same 
length,  are  equally  accurate  measurements,  although  in  one 
of  them  only  one  decimal  place  is  used,  and  in  another  five 
places. 

Measurements  which  are  expressed,  like  these,  in  three 
figures,  are  said  to  be  of  three-figure  accuracy;  they  may  be 
made  with  ordinary  instruments  and  skill.  Five-figure 
accuracy  requires  good  instruments  and  great  care,  as  where 
a  city  building  lot  is  measured  to  thousandths  of  a  foot;  six- 
figure  accuracy  is  possible  to  experts;  and  eight-figure 
accuracy  is  beyond  human  skill  at  present. 

In  solving  examples  with  decimal  data,  it  is  impossible 
for  the  result  to  have  any  greater  accuracy  than  the  given 
quantity.  If  the  measured  quantity  is  expressed  in  three 
figures,  it  is  useless  to  express  the  result  in  more  than  three 
figures;  hence,  any  answer  obtained  from  decimal  data 
must  be  cut  down  to  the  number  of  figures  of  the  given  data. 
For  example,  1.39  f1^.  =  1.39  X  12  in.  =  16.68  in.  =  16.7  in. 

EXERCISE  4 

1.  A  flagpole  is  made  in  two  parts,  a  heavy  base  and  a 
lighter  upper  portion;  the  entire  height  is  75  feet.  The 
height  of  the  base  bears  to  the  height  of  the  upper  portion 
the  ratio  2.    How  long  is  each  portion? 


MEASUREMENT  OF  LENGTH  19 

2.  A  large  factory  has  length  and  width  m  the  ratio  3. 
A  belt  course  of  ornamental  terra  cotta  around  the  building 
measures  640  feet.     What  is  the  length  of  the  building? 

3.  A  life-saving  crew  had  two  coils  of  rope  whose  lengths 
were  in  the  ratio  2,  and  neither  one  alone  would  reach  a 
wreck  offshore.  By  uniting  them  end  to  end,  the  entire, 
length  of  480  yards  carried  over  the  ship.  How  much  rope 
was  there  in  each  coil? 

4.  Spur  tracks  were  run  from  the  main  line  of  a  railroad 
to  two  factories,  and  required  the  building  of  1800  feet  of 
track.  The  lengths  of  the  two  lines  of  track  were  in  the 
ratio  5.     How  long  was  each? 

6.  A  piece  of  ground  350  feet  long  was  divided  in  the 
ratio  6  for  a  factory  and  an  adjoining  power  house.  How 
long  was  the  power  house? 

6.  A  line  87.32  feet  long  is  divided  into  two  parts  that 
have  a  ratio  7.     How  long  is  each  part? 

7.  The  length  of  a  building  lot  is  twice  its  breadth,  and 
it  takes  248.4  feet  of  iron  fence  to  inclose  it.  What  are  the 
dimensions  of  the  lot? 

8.  The  widths  of  two  adjoining  estates  have  a  ratio  13, 
and  they  occupy  a  total  frontage  of  493.5  ft.  How  wide  is 
each  estate? 

9.  The  width  of  the  wagonway  in  a  certain  street  bears 
to  the  width  occupied  by  both  sidewalks  the  ratio  2.  The 
total  width  from  wall  to  wall  is  97.29  feet.  How  wide  is 
the  wagonway? 

10.  Two  sides  of  a  triangle  have  the  ratio  3,  and  the  third 
side  is  15.8  inches  long.  The  total  perimeter  (the  distance 
around  the  edge)  is  36.7  inches.     How  long  is  each  side? 

11.  A  metal  scale  .389  ft.  long  is  divided  into  two  parts 
with  the  ratio  2.    Find  the  length  of  each  part  in  inches. 


20  THE  FIRST  USE  OF  ALGEBRA 

12.  A  meter  is  39.37  inches.  A  meter  stick  is  divided  into 
two  parts  whose  ratio  is  8.    Find  the  length  of  each  part  in  feet. 

13.  A  Hter  of  dry  air  at  standard  pressure  and  temper- 
ature weighs  1.293  grams.  What  is  the  weight  of  the  two 
portions  of  this  air  when  it  is  divided  in  the  ratio  6? 

14.  A  short  piece  of  railroad  is  laid  with  heavy  rails 
weighing  90  lb.  to  the  foot,  and  lighter  rails  weighing  75  lb. 
to  the  foot.  The  length  of  the  road  is  5.79  miles,  and  the 
distance  laid  with  heavy  rails  bears  to  the  distance  laid  with 
the  lighter  rails  the  ratio  3.  How  many  rods  of  each  grade 
of  rail  are  laid?  How  many  tons  of  each  grade  of  rail  are 
used?     (Two  rails  to  the  track.) 

16.  The  chemical  elements  in  anthracite  coal  are  mainly 
carbon,  hydrogen,  and  oxygen,  with  the  weight  of  carbon 
bearing  a  ratio  of  38  to  each  of  the  other  two.  In  |  of  a  ton 
of  coal,  how  many  pounds  of  each  of  these  constituents 
would  be  found? 

16.  A  kilogram  (1000  grams)  is  equivalent  to  2.20  lb. 
avoirdupois.  Eleven  kilograms  of  copper  is  divided  into  two 
masses  with  a  ratio  6.  How  many  ounces  are  there  in  each 
portion? 

17.  Two  distances  have  a  ratio  5  and  their  sum  is  .139 
miles.     How  many  rods  are  there  in  each  distance? 

18.  I  wish  to  divide  3.37  lb.  of  sulphur  into  two  portions 
having  a  ratio  3.  How  many  ounces  will  there  be  in  each 
portion? 

19.  A  piece  of  ribbon  2.46  yards  long  was  divided  into 
two  portions  whose  ratio  was  4.  How  many  inches  were 
there  in  each  portion? 

20.  A  book  consisting  of  two  parts  has  in  Part  II  three 
times  as  many  pages  as  in  Part  I.  The  book  has  1436  pages, 
and  the  thickness  between  covers  is  2.59  in.  How  many 
pages  are  there  in  each  part?    How  thick  is  one  page? 


MEASUREMENT  OF  ANGLES  21 

EXERCISE  6 

Oral  Work 

Find  the  lengths  of  the  pairs  of  lines  described  in  the  fol- 
lowing examples: 

1.  Sum,  12  inches;  ratio,  3. 

2.  Sum,  40  centimeters;  ratio,  4. 

3.  Sum,  75  meters;  ratio,  2. 

4.  Sum,  90  inches;  ratio,  8. 

5.  Sum,  144  feet;  ratio,  11. 

6.  Sum,  150  yards;  ratio,  5. 

7.  Sum,  55  miles;  ratio,  10. 

8.  Sum,  3  feet;  ratio,  5. 
G.   Sum,  2  inches;  ratio,  7. 

10.  Sum,  4  yards;  ratio,  11. 

11.  Sum,  300  feet;  ratio,  5. 

12.  Sum,  360  yards;  ratio,  8. 

13.  Sum,  400  feet;  ratio,  9. 

14.  Sum,  1200  miles;  ratio,  3. 

15.  Sum,  880  yards;  ratio,  7. 

16.  Sum,  960  kilometers;  ratio,  2. 

17.  Sum,  540  feet;  ratio,  8. 

18.  Sum,  1400  miles;  ratio,  13. 

19.  Sum,  484  yards;  ratio,  3. 

20.  Sum,  1111  feet;  ratio,  10. 

Measurement  of  Angles 

16.  An  angle  is  measured  by  marking  off  on  it  a  standard 
angle  called  a  degree. 

17.  If  on  a  flat  surface  you  revolve  a  straight  line  about 
one  of  its  ends  as  a  pivot,  you  get  one  complete  rotation 
when  it  returns  to  its  starting  position,  and  you  have  passed 


22 


THE  FIRST  USE  OF  ALGEBRA 


the  line  through  360  of  the  standard  angles  or  degrees.  A 
clock  hand,  or  a  watch  hand,  makes  such  a  movement,  and 
the  length  of  the  hand  has  nothing  to  do  with  the  size  of  the 
angle  through  which  it  rotates.  In  a  quarter  of  an  hour 
every  minute  hand  describes,  or  passes  over,  an  angle  of  90°, 
and  every  hour  hand  describes  an  angle  of  7J°,  whether  the 
hands  are  on  a  lady's  watch  or  on  the  town  clock.  The  size 
of  the  angle  has  nothing  to  do  with  the  length  of  its  sides. 


90 


260       27 


80 

mi 


'"%,. 

%" 


0       280 


EmtractoE 


18.  A  degree  is  divided  into  sixtieths,  called  minutes;  and 
then  subdivided  into  sixtieths  of  a  minute,  called  seconds. 
Subdivisions  of  a  degree  are  expressed  also  in  the  form  of 
decimals;  that  method  will  be  generally  used  in  this  book. 


MEASUREMENT  OF  ANGLES 


23 


19.  The  point  from  which  the  two  sides  of  an  angle  are 
drawn  is  called  the  vertex. 

The  most  convenient  way  of  measuring  an  angle  is  to 
place  it  with  its  vertex  at  the  center  of  a  circle  of  which  the 
edge  is  marked  off  in  degrees.  Such  a  marked  circle  is 
called  a  protractor. 

On  a  large  circle  the  degree 
marks  are  farther  apart  than 
on  a  small  circle,  just  as  a 
large  clockface  has  longer 
minute  spaces  (6°  each)  than 
a  small  one;  but  the  central 
angle  of  one  degree  will,  of 
course,  be  the  same,  whatever 
the  size  of  the  protractor.  

20.  Two  angles  are  called  complements  when  their  sum 
is  90°.  Complements  intercept  a  quadrant  on  a  protractor. 
Two  angles  are  called  supplements  when  their  sum  is  180°. 
Supplements  intercept  a  semicircle  on  a  protractor. 


—  Semicircle  =iz 

21.  For  representing  in  algebra  the  length-number  of  a 
straight  line,  it  is  advisable  to  use  one  of  the  small  letters  of 
the  alphabet.  For  the  number  of  degrees  in  an  angle,  use 
a  capital  letter. 


24  THE  FIRST  USE  OF  ALGEBRA 

EXERCISE  6 

1.  Two  supplementary  angles  are  in  the  ratio  8;  find 
the  angles. 

a.  An  angle  of  169°  is  divided  into  two  parts  with  the 
ratio  12.     How  many  degrees  are  there  in  each  part? 

3.  Two  angles  are  in  the  ratio  7,  and  their  sum  is  152°. 
What  is  the  size  of  each  angle? 

4.  One  angle  is  twice  the  size  of  another  angle,  and  their 
sum  is  the  supplement  of  three  times  the  larger  one.  Find 
the  two  angles. 

5.  An  angle  is  the  complement  of  5  times  itself;  find 
the  angle. 

6.  The  ratio  of  two  complementary  angles  is  3;  what 
are  the  angles? 

•7.  An  angle  of  104°  is  divided  into  two  parts,  having  a 
ratio  7.     How  many  degrees  are  there  in  each  part? 

8.  Two  supplementary  angles  have  a  ratio  17;  what  are 
the  angles? 

9.  An  angle  bears  to  its  complement  the  ratio  9;  how 
many  degrees  are  there  in  the  angle? 

10.  A  mark  on  a  clockface  is  so  placed  that  at  4  o'clock  it 
divides  the  angle  between  the  hands  into  two  angles  that 
have  the  ratio  4.  How  many  degrees  are  there  in  each  of 
these  two  angles? 

EXERCISE  7 
Oral  Work 

1.  Find  two  complementary  angles  whose  ratio  is 

1;  2;  3;  4;  5;  8;  9;  11;  14;  19;  29;  44;  f ;  f;  f ;  f ;  ^;  /^. 

2.  Find  two  supplementary  angles  whose  ratio  is 

2;  4;  5;  8;  9;  11;  14;  17;  19;  29;  44;  59;  89;  179;  |;  f ; 


MEASUREMENT  OF  ANGLES  25 

Find  the  pairs  of  angles  described  as  follows: 

3.  Sum,  77°;  ratio,  6.  9.   Difference,  10°;  ratio,  6. 

4.  Sum,  63°;  ratio,  8.  lo.   Difference,  36°;  ratio,  10. 
6.   Sum,  105°;  ratio,  20.        ii.   Difference,  80°;  ratio,  9. 

6.  Sum,  105°;  ratio,  6.  12.   Difference,  80°;  ratio,  17. 

7.  Sum,  105°;  ratio,  14.         13.   Difference,  30°;  ratio,  4. 

8.  Sum,  112°;  ratio,  27.         14.   Difference,  78°;  ratio,  14. 

22.  Where  the  number  of  degrees  in  an  angle  is  represented 
by  an  algebraic  expression,  it  is  sometimes  necessary  to  get 
another  algebraic  expression  to  represent  the  number  of 
degrees  in  the  supplement  or  the  complement. 

Let  us  take  numbers  first,  as  an  illustration,  and  find  how 
many  degrees  there  are  in  the  supplement  of  an  angle  of  133°. 

Since  47^*  +  133°  =  180°,  47°  is  the  supplement  of  133°  (that  is,  of 
180°  -  47°). 

Likewise,  133°  is  the  supplement  of  47°  (that  is,  of  180°  -  133°). 

Hence,  the  number  of  degrees  in  the  supplement  of  7°  is  180  —  7;  of 
(c  +  d)  degrees  is  (180  —  c  —  d). 

The  number  of  degrees  in  the  complement  of  (a  +  4)  degrees  is 
(90  -  a  -  4),  that  is,  (86  -  a). 

Each  of  the  following  algebraic  expressions  represents  the 
number  of  degrees  in  an  angle;  state  in  each  case  the  number 
of  degrees  in  its  supplement. 

15.  M       18.   ah  21.   c  +  8         24.   2a +  3 

16.  K        19.   r  +  s  22.   c  +  24       25.   6  R 

17.  Q         20.    Z+F  23.   /+11         26.    2/1  +  15 

Each  of  the  following  algebraic  expressions  represents  the 
number  of  degrees  in  an  angle;  state  in  each  case  the  number 
of  degrees  in  its  complement. 

27.  Z        30.   ef  33.   ikf  +  1       36.   6  a  +  7 

28.  S        31.  h-\-k  34.  c  +  9         37.   3cd 

29.  g        32.  MN  35.   11  + It;      38.  m  +  5a  +  4 


26  THE  FIRST  USE  OF  ALGEBRA 

Valuation  Problems 

23.  Model  D.  —  I  settled  an  account  of  $48  with  $2 
and  $5  bills,  using  twice  as  many  5's  as  2's.  How  many  bills 
of  each  denomination  did  I  give  in  payment? 

Let  X  =  the  number  of  $2  bills. 

Then    2  a;  =  the  number  of  $5  bills. 

2x  =  the  number  of  dollars  in  $2  bills. 
10  X  =  the  number  of  dollars  in  $5  bills. 
®      10  X  +    2  a;  =  48 
@  12  a;  =  48         ®  same  values 

®  a;  =  4  @  ^  12 

0  2a;  =  8  ®  X  2 

Ans.    Four  $2  bills;  eight  $5  bills. 
Check:  4  $2  bills  =  $  8 

2  X  4  or  8  $5  bills  =    40 
$48 

EXERCISE  8 

1.  The  coal  supply  for  a  school  was  delivered  in  two  kinds 
of  wagons,  the  larger  holding  5i  tons  and  the  smaller  3  tons; 
there  were  twice  as  many  of  the  large  loads  as  of  the  small 
ones,  and  the  entire  amount  deUvered  was  182  tons.  How 
many  loads  were  delivered  by  each  kind  of  wagon? 

2.  A  milk  wagon  carries  certified  milk  in  bottles  at  14 
cents  a  quart  and  5  times  as  much  milk  in  cans  at  9  cents  a 
quart.  The  entire  load  is  worth  $16.52.  How  many  quarts 
of  each  kind  of  milk  are  carried  by  the  wagon? 

3.  A  stock  farm  sold  out  its  horses  and  cows  for  $4360. 
There  were  three  times  as  many  cows  as  horses.  The  cows 
were  sold  at  a  uniform  price  of  $110,  and  the  horses  at  a 
uniform  price  of  $215.     How  many  of  each  were  sold? 

4.  Two  kinds  of  wagons,  one  kind  holding  twice  as  much 
as  the  other,  were  used  in  delivering  220  tons  of  brick.  The 
smaller  wagons  delivered  16  loads,  the  larger  31.  How 
many  tons  were  delivered  in  each  kind  of  wagon? 


VALUATION  PROBLEMS  27 

6.  A  shoe  dealer  sold  one  day  a  line  of  shoes  at  $6  a  pair, 
and  four  times  as  many  of  another  line  at  $3.50  a  pair. 
The  receipts  on  these  two  lines  of  shoes  that  day  were  $460. 
How  many  pairs  of  each  kind  were  sold? 

6.  In  paying  an  account  of  $68,  I  used  only  $5  and  $2 
bills;  and  the  number  of  $5  bills  was  three  times  the  number 
of  $2  bills.     Find  the  number  of  bills  of  each  denomination. 

How  many  bills  of  each  kind  must  I  use  to  settle  the  fol- 
lowing accounts  in  the  manner  described  for  each  account? 

7.  To  pay  $88,  I  use  $5  and  $2  bills  only,  3  times  as  many 
2's  as  5's. 

8.  To  pay  $72,  I  use  5's  and  2's  only,  twice  as  many 
5's  as  2's. 

9.  To  pay  $42,  I  use  lO's  and  2's  only,  twice  as  many  2's 
as  lO's. 

10.  To  pay  $80,  I  use  lO's  and  5's  only,  twice  as  many 
5's  as  lO's. 

11.  To  pay  $26,  I  use  lO's  and  I's  only,  3  times  as  many 
I's  as  lO's. 

12.  To  pay  $45,  I  use  2's  and  I's  only,  7  times  as  many 
2's  as  I's. 

is.  To  pay  $78,  I  use  5's  and  I's  only,  5  times  as  many 
5's  as  I's. 

14.  To  pay  $84,  I  use  lO's  and  2's  only,  4  times  as  many 
lO's  as  2's. 

15.  To  pay  $75,  I  use  lO's  and  5's  only,  7  times  as  many 
lO's  as  5's. 

16.  To  pay  $183,  I  use  lO's  and  I's  only,  6  times  as  many 
lO's  as  I's. 

17.  Five  horses  and  four  donkeys  weigh  altogether  9600 
pounds.  Each  horse  weighs  4  times  as  much  as  a  donkey. 
Find  the  weight  of  each  animal. 


28  THE  FIRST  USE  OF  ALGEBRA 

18.  Apples  were  bought  for  39  cents;  twice  as  many  red 
ones  at  5  cents  apiece  as  green  ones  at  3  cents  apiece.  How 
many  of  each  kind  of  apples  were  bought? 

19.  Engineers  examined  a  bridge  that  broke  down  under 
the  weight  of  a  crowd  of  people,  and  decided  that  the  break- 
ing load  was  36  tons.  Supposing  that  there  were  four  times 
as  many  men  as  women,  the  average  weight  being  150  pounds 
for  a  man  and  120  pounds  for  a  woman,  what  was  the  num- 
bei*  of  each  in  that  crowd? 

20.  I  bought  a  certain  quantity  of  tea  of  first  quality  at 
60  cents  a  pound,  and  3  times  as  much  tea  of  second  quality 
at  45  cents  a  pound.  The  tea  cost  me  $13.65.  How  many 
pounds  of  each  quality  did  I  buy? 

21.  Mules  at  $40  apiece  and  7  times  as  many  horses  at 
$125  apiece  cost  $2745.  Find  the  number  of  animals  in  the 
drove. 

22.  Lead  weighs  4  times  as  much  as  marble;  3  lead 
globes  and  2  marble  globes,  all  of  the  same  size,  weigh  56 
pounds.     What  is  the  weight  of  each  globe? 

23.  A  team  is  made  up  of  oxen  and  mules,  one  ox  and 
one  mule  in  each  pair;  supposing  that  an  ox  can  pull  4 
times  as  much  as  a  mule,  and  that  the  whole  team  is  used  to 
pull  30  tons,  how  much  of  this  load  do  the  oxen  pull? 

24.  I  paid  $91  with  $5  bills  and  $2  bills,  using  the  same 
number  of  each.     How  many  of  each  did  I  use? 

26.  There  are  three  times  as  many  5's  as  2's  in  a  roll  of 
bills;  in  all  $51.     Find  the  number  of  bills  of  each  kind. 

26.  I  paid  a  bill  of  $1.80  with  quarters,  dimes,  and  nickels, 
using  twice  as  many  dimes,  and  3  times  as  many  nickels,  as 
quarters.     Find  the  number  of  each  that  I  used. 

27.  I  bought  twice  as  much  coffee  at  33  cents  a  pound  as 
tea  at  54  cents,  and  I  paid  in  all  $2.40.  How  many  pounds 
of  each  did  I  buy? 


VALUATION  PROBLEMS  29 

28.  I  bought  horses  at  $120,  sheep  at  $14,  and  chickens 
at  50  cents;  three  times  as  many  sheep  as  horses,  seven 
times  as  many  chickens  as  sheep.  All  cost  $517.50.  How 
many  of  each  did  I  buy? 

29.  The  crew  of  a  towboat  consists  of  an  engineer,  2 
firemen,  3  deckhands,  and  a  cook,  whose  wages  are,  respec- 
tively, $2,  $1.50,  $1,  and  $1.20  per  day.  For  a  certain 
voyage  the  pay-roll  was  $73.60.  How  many  days  did  the 
voyage  last? 

30.  A  certain  mill  employs  men  at  an  average  wage  of 
$2.34  a  day,  5  times  as  many  women  at  an  average  wage  of 
$1.26  a  day,  and  twice  as  many  children  (as  men)  at  an 
average  wage  of  62  cents  a  day.  The  weekly  pay-roll  of 
this  mill  is  $592.80.  How  many  men,  women,  and  children 
are  employed  in  it? 

31.  I  have  three  times  as  many  nickels  as  half-dollars;  in 
all  $4.55.     Find  the  number  of  each. 

32.  There  are  in  a  purse  three  times  as  many  nickels  as 
dimes;   in  all  $1.50.     How  many  nickels  are  there? 

33.  Four  cows,  three  calves,  and  ten  sheep  cost  $168;  a 
cow  costs  five  times  as  much  as  a  calf,  and  a  calf  costs  twice 
as  much  as  a  sheep.     Find  the  cost  of  each. 

34.  The  ratio  of  two  angles  is  6.  Double  the  smaller 
one  is  the  complement  of  the  larger.     Find  the  angles. 

35.  The  difference  of  two  lengths  is  32  feet  and  their 
ratio  is  5.     Find  the  lengths. 

36.  The  ratio  of  two  angles  is  6,  and  the  double  of  the 
smaller  is  the  supplement  of  three  times  the  larger.  Find 
the  angles. 

37.  Five  wooden  posts  and  two  iron  ones  of  exactly  the 
same  size  and  shape  weigh  altogether  810  pounds.  The  ratio 
of  the  weight  of  iron  to  the  weight  of  this  wood  is  11.  Find 
the  weight  of  each  post. 


30  THE  FIRST  USE  OF  ALGEBRA 

Difference  of  the  Unknown  Numbers  Given 

24.  Model  E.  —  A  father  is  50  pounds  heavier  than  his 
son,  and  both  weigh  248  pounds.  What  is  the  weight  of 
each? 

Let  X  =  the  number  of  pounds  the  son  weighs. 

Then  x  +  50  =  the  number  of  pounds  the  father  weighs. 


©              X 

+  X  +  50 

=  248 

® 

2X  +  50 

=  248 

© 

same  values 

® 

2x 

=  198 

© 

-50 

® 

X 

=    99 

© 

^    2 

© 

x  +  50 

=  149 

©  +  50 

Ans. 

Father, 

149 

pounds; 

son, 

99 

pounds, 

99  +  50 

=    149 

+  99 

248 

EXERCISE  9 

Check; 


1.  Two  cannon  balls,  one  being  6  pounds  heavier  than 
the  other,  together  weighed  50  pounds.  What  was  the 
weight  of  each  cannon  ball? 

2.  I  have  ten  cents  more  than  my  uncle,  and  together  we 
have  $2.90.    How  much  has  each? 

3.  A  marketman  sold  11  sheepskins,  and  then  lost  $2 
of  the  money  he  received;  he  had  $7.90  left.  For  how  much 
did  one  sheepskin  sell? 

4.  A  man  owes  $181;  he  pays  part  of  the  account  with 
equal  numbers  of  $10,  $5,  and  $2  bills,  and  gives  a  check  for 
$28  for  the  remainder.  How  many  bills  of  each  kind  does 
he  give? 

5.  Find  four  consecutive  numbers  whose  sum  is  94. 

6.  Two  mules  and  a  horse  were  bought  for  $400,  and  the 
horse  cost  $40  more  than  a  mule.    What  was  the  cost  of  each? 

7.  Two  bucketfuls  of  water  fill  a  15-gallon  tub  all  but  5 
quarts.    How  much  does  the  bucket  hold? 


DIFFERENCE  OF  THE  UNKNOWN  NUMBERS  GIVEN     31 

8.  A  grocer  mixes  tea  that  costs  him  20  cents  a  pound 
with  3  times  as  much  tea  that  costs  him  38  cents  a  pound, 
and  sells  the  mixture  for  $10,  clearing  a  profit  of  62  cents. 
How  many  pounds  of  each  kind  of  tea  does  he  use? 

9.  My  farm  is  three  times  as  large  as  the  one  next  to  it, 
and  both  together,  with  a  10-acre  lot  across  the  road,  just 
make  a  quarter  section  (160  acres).  Find  the  size  of  each 
farm. 

10.  There  are  quarters  in  one  pile,  twice  as  many  dimes 
in  another,  and  $1.13  besides;  all  this  money  amounts  to 
$6.08.    How  many  quarters  and  dimes  are  there? 

/^ii.  Thirty  horses  and  40  mules  weigh  54  tons;  on  an 
average,  each  horse  weighs  100  pounds  more  than  a  mule. 
Find  the  average  weight  of  each  kind  of  animal. 

12.  I  bought  peas  at  59  cents  a  peck,  and  1  peck  more 
of  beans  at  30  cents  a  peck;  also  25  cents  worth  of  lettuce. 
I  paid  in  all  $5.  How  many  pecks  of  peas  and  of  beans  did 
I  buy? 

13.  I  bought  sugar  at  5  cents  a  pound,  3  times  as  much 
coffee  at  50  cents  a  pound,  and  paid  $1.83  for  a  ham.  The 
whole  bill  was  $12.68.     How  much  coffee  did  I  buy? 

14.  I  sold  a  house;  then  I  sold  5  more  houses  at  double 
the  price;  then  sold  35  tons  of  coal  at  $5  per  ton.  I  re- 
ceived in  all  $64,448.*  What  was  the  price  of  the  first  house 
sold? 

15.  A  man  has  three  times  as  many  dimes  as  dollars, 
4  times  as  many  cents  as  dimes,  5  times  as  many  nickels  as 
cents.     He  has  in  all  $17.68.     Find  the  number  of  nickels. 

16.  Suppose  that  each  of  the  36  boys  in  a  class  has  the 
same  sum  of  money,  and  that  the  teacher  has  15  cents  more 
than  all  the  boys  together.  All  the  money  amounts  to 
$79.35.     How  much  has  the  teacher? 


32 


THE  FIRST  USE  OF  ALGEBRA 


17.  Three  times  as  many  $10  bills  as  $2  bills,  and  $7.35 
besides,  make  a  total  of  $167.35.    Find  the  number  of  $2  bills. 

18.  Seven  times  as  many  $1  bills  as  $5  bills,  and  $1.45 
besides,  make  $37.45.     Find  the  number  of  $1  bills. 

19.  Three  times  as  many  $2  bills  as  $5  bills,  4  times  as 
many  $10  bills  as  $2  bills,  and  $11.11  besides,  make  $666.11. 
Find  the  number  of  $2  bills 

Angles  and  Angle  Relations 

25.  Two  lines  are  called  parallel  when  they  are  on  the 
same  flat  surface  and  do  not  meet,  no  matter  how  far  they 
may  be  prolonged  in  either  direction. 

The  figure  made  by  two  parallel  lines  is  called  a  stripe. 

A  straight  Hne  that  crosses  two  or  more  straight  lines  is 
called  a  transversal. 


Fig.  2. 


26.  When  a  transversal  crosses  a  stripe,  it  makes  two 
sets  of  four  angles  each.  If  we  number  each  set  with 
suffixes,  beginning  each  time  with  the  upper  angle  on  the 
right  hand  side,  it  will  be  seen  that 

All  the  odd  angles  are  equal. 

All  the  even  angles  are  equal. 

The  odd  angles  are  the  supplements  of  the  even  angles. 

The  angle  Ai  is  read  "A  one." 


ANGLES  AND  ANGLE  RELATIONS  33 

27.  In  the  triangle  ABC  (Fig.  2),  the  sum  of  the  angles 
at  the  base  is  equal  to  the  supplement  of  the  angle  at  the 
vertex.     That  is,       A  +  C  =  180  -  B. 

Prolong  CB,  and  draw  through  B  a  Une  parallel  to  CA.     We  now 
have  one  stripe  with  two  transversals.     On  the  transversal  AB,  A  and 
X  are  even  angles  and  are  equal.     On  the  transversal  CB,  C  and  Y  are 
odd  angles  and  are  equal.     That  is, 
©  A  =X 

®  C  =  Y 

But         0         X  +  F  =  180  -  B 
Hence,    ®  A+C  =  180-B     Substituting  ®  and  ®  in  ® 

Since  this  equation  may  be  written 

®       A+5  +  C  =  180        0  +  5 
we  have  the  following  theorem:* 

The  sum  of  the  angles  of  a  triangle  is  180°. 

28.  Model  F.  —  In  a  certain  triangle,  the  two  base  angles 
have    a    ratio    2,    and    the 

ratio  of   the  vertical   angle 

to  the  larger  base  angle  is 

3.    Find  the  angles  of  the 
triangle. 

Let         C  =  the  number  of  degrees  in  the  smaller  base  angle. 

Then   2  C  =  the  number  of  degrees  in  the  larger  base  angle. 

and  6  C  =  the  number  of  degrees  in  the  vertical  angle. 

0       C  +  2C  +  6C  =  180 

0  9  C  =  180        0  same  values 

0  C  =    20        0  --  9 

0  2C=40        0X2 

0  6C  =  12O0X3 

Am.    20°,  40°,  120°. 
Check:  20 

2  X  20  =    40 

3  X  40  =  120 

180 

*  A  theorem  is  any  general  statement  for  which  we  may  expect  to 
find  a  proof. 


34  THE  FIRST  USE  OF  ALGEBRA 

EXERCISE  10 

1.  In  the  figure  made  by  a  stripe  and  its  transversal,  the 
odd  and  even  angles  have  a  ratio  3.  How  many  degrees  are 
there  in  each  angle? 

2.  In  the  same  kind  of  figure,  each  obtuse  angle  is  24° 
greater  than  any  acute  angle.  What  are  the  interior  angles 
on  one  side  of  the  transversal? 

3.   In  the  stripe  shown  in  Fig.  1, 
the    angle  A   is   21°  greater   than 
-  double  the  angle  B.     How   many 
degrees  are  there  in  each? 

4.   In  the  same   kind  of  figure, 

^  angle  A    is    5°   more    than    three 

^  times    angle    B.      Find    the    two 

^«-  1-      '  angles. 

5.  In  the  same  kind  of  figure,  angle  A  is  13.62°  more  than 
twice  angle  B.     How  many  degrees  are  there  in  each  angle? 

6.  In  the  same  kind  of  figure,  angle  A  is  31.13°  more  than 
four  times  angle  B.     How  many  degrees  in  each  angle? 

7.  Three  of  the  odd  angles  in  a  stripe  make  a  total  40° 
less  than  one  of  the  even  angles.  How  many  degrees  are 
there  in  each  angle? 

8.  One  angle  of  a  triangle  is  20°,  and  the  other  two  angles 
have  a  ratio  7.    What  is  the  largest  angle? 

9.  The  two  base  angles  of  a  triangle  have  a  ratio  3,  and 
the  smaller  is  15°  less  than  the  vertical  angle.  What  are 
the  angles? 

10.  The  vertical  angle  of  a  triangle  is  10°  greater  than  one. 
of  the  base  angles,  and  the  other  base  angle  is  equal  to  the 
sum  of  the  two.     Find  the  vertical  angle. 

11.  The  base  angles  of  a  triangle  are  equal,  and  their 
sum  is  18°  less  than  the  vertical  angle.  What  are  the  base 
angles? 


ANGLES  AND  ANGLE  RELATIONS 


35 


12.  In  the  triangle  ABC  (Fig.  2),  the  exterior  angle  at  A  is 
equal  to  108°,  and  the  ratio  of  5  to  C  is  2.6.  Find  the  three 
angles  of  the  triangle. 


Fig.  2. 

13.  In  the  same  kind  of  figure,  angle  A  is  5°  more  than 
angle  B  and  7°  more  than  angle  C.  How  many  degrees  are 
there  in  each  angle? 

14.  In  the  same  kind  of  figure,  angle  J.  is  3  times  angle 
B,  and  angle  C  is  30°  more  than  angle  B.  How  many  de- 
grees are  there  in  A,  B,  and  C? 

In  the  same  kind  of  figure,  angle  A  is  15.3°  more  than 


15. 


B,  and  B  is  5  times  angle  C. 
in  each? 


How  many  degrees  are  there 


Fig.  3. 

16.  In  Fig.  3,  angle  B  is  16.38°  more  than  angle  A, 
angles  A  and  B, 

17.  In  Fig.  4,  angle  C  is  20.34°  more  than  angle  D, 
many  degrees  are  there  in  the  angles  C  and  D? 


Find 


How 


36  THE  FIRST  USE  OF  ALGEBRA 

18.  In  the  triangle  MNO,  angle  M  is  30.5°  more  than 
angle  0,  and  angle  N  is  11.3°  more  than  angle  0.  Find  the 
number  of  degrees  in  each  angle. 

19.  In  triangle  ABC,  angle  A  +  angle  B  =  angle  C,  and 
A  is  32.77°  more  than  B.  Find  the  number  of  degrees  in 
the  angles  A,  B,  and  C 

20.  An  angle  A  +  an  angle  B  =  45.38°,  and  angle  A  = 
2.23  times  angle  B,     Find  angles  A  and  B. 

Let  X  =  number  of  degrees  in  angle  A;  2.23  x  =  number  of  degrees 
in  angle  B. 

21.  In  a  triangle  EFG,  E  =  F  =  1.34  G,  Find  angles 
E,  F,  and  G. 

Multiplications 

29.  Consider  the  problem:  A  is  twice  as  old  as  B;  22 
years  ago  he  was  three  times  as  old  as  B.  What  are  their 
ages  now? 

Let  X  *=   the  number  of  years  in  B's  age  now. 

22  years  ago  A  was  2  x  —  22  years  old,  and  B  was  x  —  22  years  old. 
So  the  equation  is  2  x  -  22  =  3  (x  -  22) 

Before  solving  this  equation,  we  must  understand  ex- 
pressions like  3  (a;  —  22). 

Suppose  a  farmer  contracts  to  deliver  3  bushels  of  oats 
and  2  bushels  of  barley  every  day.  To  get  the  number  of 
bushels  of  grain  delivered  in  several  days,  he  multiplies 
not  only  the  quantity  of  oats  but  also  the  quantity  of  barley 
by  the  number  of  days. 

Suppose  a  farmer  receiv.es  daily  x  bushels  of  grain  and  delivers  daily 
3  bushels.  The  increase  of  grain  in  his  store  for,  say,  7  days  is  found 
thus; 

X  bushels  daily  for  7  days  received  7  x; 

3  bushels  daily  for  7  days  delivered  21. 

Subtracting  the  amount  delivered  from  the  amount  received  (7  x  — 
21),  we  find  what  the  daily  increase  of  x  —  3  amounts  to  in  7  days. 


SHORTAGES  37 

30.   Whenever  an  expression  of  two  or  more  terms  is  mul- 
tiplied,  each    term   of  that   expression   must  he  multiplied 

separately. 

EXERCISE  11 


1. 

2  (a;  -  7)           5. 

75(2X  +  40) 

9. 

125  (8  a: +  40) 

2. 

11  (3  a;  -  1)      6. 

50  (8  a  +  160) 

10. 

13(180-7  5) 

3. 

18(3x  +  5)      7. 

20  (5  A +  70) 

11. 

30  (5 -3m) 

4. 

25(?4x+28)    8. 

40  (25  x+ 125) 

12. 

23  (23  -  x) 

Shortages 

31.  The  equation  2  x  -  22  =  3  (x  -  22)  in  Section  29 
may  now  be  changed  to  2  x  —  22  =  3  x  —  66,  which  is  a 
change  of  form  but  not  of  value.  This  equation  differs 
from  those  we  have  been  using  in  that  it  has  negative 
terms  on  each  side  of  the  equation,  that  is,  terms  to  be 
subtracted. 

The  left  side  is  not  equal  to  2  x,  but  just  22  short  of  it. 
In  the  same  way,  the  right  side  is  66  short  of  3  x. 

If  22  is  added  to  each  side  of  the  equation,  the  shortage 
on  the  left  side  will  disappear,  but  the  right  side  will  still 
be  44  short  of  3  x.     That  is,  the  equation  will  be 

2  X  =  3  X  -  44. 

This  shortage  disappears  when  44  is  added  to  each  side, 
giving  2  X  +  44  =  3  x,  an  equation  Uke  many  we  have  been 
solving.  The  values  on  each  side  have  been  changed,  but 
the  resulting  numbers  are  equal. 

But  if  66  is  added  at  first,  we  have 

2x-  22  +  66  =  3x 
2x  +  44=3x 

The  entire  shortage  on  the  right  is  made  up;  and  on  the  left 
only  22  out  of  the  66  is  needed  to  make  up  that  shortage, 
leaving  44  to  be  added  to  the  complete  value  of  2  x. 


38  THE  FIRST  USE  OF  ALGEBRA 

The  shortest  way,  then,  when  there  are  two  similar  short- 
ages, is  to  add  the  larger.  The  numbers  that  were  stated  as 
equal  in  the  equation  2a;  —  22  =3x  —  66  have  each  be- 
come greater;  they  are  now  different  in  value  from  the 
original  numbers,  but  are  still  equal  to  each  other. 

32.  Model  G.  —  A  is  twice  as  old  as  B;  22  years  ago  he 
was  three  times  as  old  as  B.    What  are  their  ages  now? 

Let  X  =  the  number  of  years  in  B's  age  now. 

Then    2  x  =  the  number  of  years  in  A's  age  now. 

a;  —  22  =  the  number  of  years  in  B's  age  22  years  ago. 
2  X  —  22  =  the  number  of  years  in  A's  age  22  years  ago. 
©       2  x  -  22  =  3  (x  -  22) 
@       2x-22  =  3a;-66         ©  same  values 
®       2x  +  44  =  3x  ®  +  66 

®  U  =     X  ®-2aj 

®  88  =  2x  ®X2 

Ans.     A  is  88  years  old;  B,  44  years  old. 
Check:        B's  age:  now,  44;  22  years  ago,  22. 

A's  age:  now,  88;  22  years  ago,  66.     66  =  3  X  22. 

EXERCISE  12 
Solve  for  the  value  of  the  letter : 

1.  5(A  -7)  =3A  -9         6.  8(X-3)  =5X-3 

2.  7(a;-3)=5x-3  7.   5(x  +  4)=7a:-4 

3.  4  (  X  +  5)  =  6  a;  -  20        8.   11  (x  +  1)  =  13  a;  +  1 

4.  3  (  X  +  2)  =  4  X  -  5  9.    13  (x  -  10)  =  3  X 

6.   6  (B  -  3)  =  5  (B  -  2)     lo.   9  (x  +  6)  =  10  (2  x  +  1) 

11.  A  is  now  5  times  as  old  as  B,  and  5  years  hence  he  will 
be  only  3  times  as  old  as  B.    What  are  their  ages  now? 

12.  A  is  3  times  as  old  as  B;  15  years  ago  he  was  5  times 
as  old  as  B.    What  are  their  ages  now? 

13.  A  is  3  times  as  old  as  B;  in  8  years  he  will  be  only 
2i  times  as  old  as  B.     Find  their  present  ages. 


SHORTAGES  39 

14.  A  is  7  times  as  old  as  B;  in  25  years  the  ratio  of  their 
ages  will  be  2.    What  are  their  ages  now? 

15.  The  difference  between  a  father's  age  and  his  son's 
age  is  28  years.  How  old  was  the  father  when  the  son  was 
i  of  the  father's  age?  ,  How  old  was  the  son  when  the  father 
was  9  times  as  old? 

16.  The  ratio  of  two  angles  is  2,  and  their  complements 
differ  by  21.33°.     Find  the  angles. 

17.  Of  the  eight  angles  formed  by  the  transversal  of  a 
stripe,  the  four  odd  ones  have  a  sum  just  12°  less  than  two 
of  the  even  ones.     Find  the  angles. 

18.  Two  angles  of  a  triangle  are  equal,  and  each  is  77.14° 
less  than  the  supplement  of  the  third  angle.  What  are  the 
angles? 

19.  There  are  four  steel  girders  to  be  used  end  to  end; 
if  they  were  each  6  feet  shorter,  they  would  just  do,  and  if 
they  were  each  6  feet  longer,  3  of  them  would  do.  How 
long  are  they? 

20.  There  are  four  iron  pipes  to  be  placed  end  to  end;  if 
they  were  each  12.4  feet  longer,  they  would  be  just  long 
enough,  and  if  they  were  each  12.4  feet  shorter,  it  would 
take  5  pipes  to  fill  the  space.     How  long  are  they? 

21.  I  paid  39  cents  with  nickels  and  cents,  using  nine 
more  cents  than  nickels.  Find  the  number  of  coins  of  each 
denomination. 

22.  I  paid  85  cents  in  nickels  and  dimes,  using  2  more  nickels 
than  dimes.    Find  the  number  of  coins  of  each  denomination. 

23.  I  paid  $2.05  in  quarters  and  nickels,  using  1  fewer 
nickels  than  quarters.  Find  the  number  of  coins  of  each 
denomination. 

24.  I  paid  $4.75  in  quarters  and  half-dollars,  using  5 
fewer  quarters  than  half-dollars.  Find  the  number  of  coins 
of  each  denomination. 


40  THE  FIRST  USE  OF  ALGEBRA 

25.  I  paid  $1.45  in  quarters  and  dimes,  using  4  more 
dimes  than  quarters.  Find  the  number  of  coins  of  each 
denomination. 

26.  I  borrowed  a  certain  number  of  quarters,  and  paid 
back  the  debt  with  4  fewer  half-dollars.  I  repaid  25  cents 
too  much.     How  many  coins  did  I  borrow? 

27.  I  borrowed  some  coins,  and  paid  back  5  fewer  coins 
of  another  sort;  the  coins  borrowed  were  quarters,  those 
repaid  were  halves.  I  still  owe  50  cents.  How  many  coins 
did  I  borrow? 

28.  Rice  costs  2  cents  a  pound  more  than  sugar;  3  pounds 
of  sugar  and  10  pounds  of  rice  come  to  $1.24.  Find  the  cost 
of  each  per  pound. 

29.  I  walk  3  miles  more  in  the  forenoon  than  in  the 
afternoon;  and  between  Monday  noon  and  Friday  night  I 
walk  75  miles.     How  far  do  I  walk  in  one  afternoon? 

30.  A  street  680  feet  long  has  on  one  side  10  house  lots; 
3  are  narrow  lots,  all  of  the  same  width;  the  other  lots  are 
each  20  feet  wider  than  the  first  lots.  What  is  the  width  of 
each  of  the  lots? 

31.  A  wall  13i  feet  high  is  built  with  7  courses  of  founda- 
tion stone  and  41  courses  of  brick;  each  course  of  stone  is 
9  inches  thicker  than  one  of  brick.  What  is  the  thickness 
of  the  foundation  stones? 

32.  A  wall  is  laid  with  bricks  of  two  thicknesses,  3  inches 
and  5  inches;  there  are  3  more  courses  of  the  thicker  kind 
than  of  the  thinner,  and  the  wall  is  9  feet,  3  inches  high. 
Find  the  number  of  courses  of  each  kind  of  brick. 

33.  A  hallway  is  laid  with  rows  of  tiles  of  two  widths, 
2§  inches  and  3§  inches.  There  are  4  more  rows  of  the 
wider  kind  of  tile;  and  the  hallway  is  5  feet,  2  inches  wide. 
Find  the  number  of  rows  of  each  kind  of  tile. 


THE  STUDY  OF  A  SIMPLE  EQUATION  41 

The  Study  of  a  Simple  Equation 

33.  In  either  member  of  an  equation,  or  in  both,  we  may 
find  terms  with  letters  only,  terms  with  figures  only,  or  terms 
with  both.  Terms  that  have  the  same  letters  are  called 
similar  terms;  the  terms  here  that  have  no  letters  are  also 
called  similar  terms.  Thus,  in  the  expression  72  x  +  45  + 
5  X  —  15,  the  terms  72  x  and  +5  x  are  similar;  and  the  terms 
+  45  and  — 15  are  similar. 

Similar  terms  can  be  united.  Thus  72  x  +  5  x  is  the 
same  value  as  77  x.  On  the  other  hand,  terms  that  are  not 
similar  cannot  be  united.  72  x  +  45  is  not  the  same  as 
117  x;  it  is  not  the  same  as  117;  it  cannot  be  simplified. 

34.  In  any  case  where  a  number  may  be  separated  into 
two  factors,  either  of  these  factors  is  called  the  coefficient 
of  the  other.  Thus  the  term  34  x  may  be  separated  into 
two  factors  in  several  ways;  for  example,  it  is  the  same  as 
17  X  2  X,  and  consequently  17  is  the  coefficient  of  2  x  in 
this  term,  and  2  x  is  the  coefficient  of  17. 

In  any  term,  the  factor  that  is  expressed  in  figures  is  of 
especial  importance;  we  call  it  the  coefficient  of  the  term. 
Thus  the  coefficient  of  the  term  34  x  is  34. 

35.  An  axiom  is  a  general  statement  not  requiring 
proof.  The  following  axioms  are  used  as  authority  for  any 
changes  in  value  that  we  make  in  the  members  of  any  simple 
equation. 

36.  //  equal  numbers  are  increased  or  decreased  by  the  same  Cv^ 
amount,  the  resulting  numbers  are  equal.  "^--^^ 

37.  If  equal  numbers  are  multiplied  or  divided  by  the  same 
amount,  the  resulting  numbers  are  equal. 

38.  These  axioms  specify  four  different  things,  —  ad- 
dition, subtraction,  multiplication,  and  division,  —  that  can 
be  done  to   the  numbers  on  each   side  of  an   equation. 


42  THE  FIRST  USE  OF  ALGEBRA 

There  are  other  things,  such  as  uniting  similar  terms,  and 
performing  indicated  multipHcations,  that  do  not  change  the 
value  of  the  equal  numbers  at  all,  though  they  do  change 
the  form  of  the  algebraic  expressions. 

In  order  to  avoid  confusion,  the  student  is  advised  to 
apply  these  changes  systematically  in  the  order  indicated  by 
the  following 

Rule 

(1)  Perform  any  indicated  multiplications. 

(2)  Unite  similar  terms  in  each  member. 

(3)  If  there  are  any  terms  with  minus  signs  before  them, 
add  to  each  member  enough  to  make  up  these  shortages. 

(4)  Subtract  from  each  member  the  smaller  unknown  term. 

(5)  Subtract  from  each  member  any  known  term  that 
stands  beside  an  unknown  term. 

(6)  Divide  each  member  by  the  coefficient  of  the  unknown 
term. 

This  rule  has  interesting  associations  with  the  history  of 
our  subject.  There  were,  in  ancient  times,  two  great  text- 
books of  algebra:  the  first,  written  in  Greek  about  1600 
years  ago,  gave  the  substance  of  the  rule;  the  second, 
written  in  Arabic  more  than  1000  years  ago,  had  for  its  title 
the  name  by  which  this  rule  was  then  known.  The  first 
word  of  that  title  is  our  modern  word  algebra;  it  means 
''restoration"  and  signifies  the  operation  called  for  in 
paragraph  (3)  of  the  rule. 

EXERCISE  13 
Unite  similar  terms: 

1.  3x- 5  +  4X  +  2  -  5a:  +  7-a; 

2.  7-7x  +  3-x-ll  +  9a:  +  a;-l 

3.  3  +  5x-8a;-2  +  7a;-10  +  x+15 

4.  13x-2x+17-5x-ll-a;  +  7x-23 
6.   H-a;  +  6-3a;- 10-|-7a;-5a;  +  4 


\ 


THE  STUDY  OF  A  SIMPLE  EQUATION  43 

Perform  indicated  multiplications  and  unite  similar  terms: 

6.  3  (4  -  x)  +  7  (x  -  3)  +  5x  -  4 

7.  5  (1  -  a:)  +  13  (x  -  3)  +  11  a:  -  19 

8.  x  +  5  +  3(2x- 1) +  7  +  2(1 -5x)  +  10a; 

9.  3  +  4(a;-5)+3x  +  7+13(4-3x)  +  ll 
10.  4  (x  -  5)  +  9  (3x  -  2)  +  X  +  4  (1  -  5a;) 

Solve  the  following  equations  for  x: 

li.  3  (x  -  7)  =  2x  -  11         16.   7  (x  -  2)  =  4  (x  +  4) 

12.  2  (x  +  5)  =  lOx  -  6         17.  X  +  6  =  7  (10  -  x) 

13.  2x  +  5=3x  18.   8x-l=5a:  +  41 

14.  5  x  -  7  =  3  (x+1)  -  4      19.   7  (x  +  2)  =  X  +  32 

15.  lOx  +  13  =  3  (x  +  9)      20.   5  (1  -  x)  =  1  -  3a; 

21.  11  (x  +  3)  =  10  (x  +  1)  +  1 

22.  8  (x  +  10)  =  12  (x  +  7)  -  8 

23.  100  (x  -  5)  =  60  (x  +  3) 

24.  4  (x  +  5)  =  2  (3  X  -  10) 

25.  3  (x  -  10)  +  2  (x  +  4)  =  6x  -  22 
26.*  8  (10  -  x)  =  5  (x  +  3) 

27.  3  X  +  4  (x  -  2)  =  1  +  3  (2  X  -  3) 

28.  19  (x  -  2)  =  6  (3  X  -  5) 

29.  3  (3  X  -  5)  =  4  (x  +  5) 

30.  15(5x-3)  =  12(4x  +  3) 

31.  5(x  +  4)  =  14x  +  2 

32.  3  (x  +  3)  =  3  (3  X  +  1)  -  12 

33.  5  (x  +  1)  =  6  (3  -  x)  -  2 

34.  2  X  =  5  (x  -  1)  -  2  X 

35.  17  (x  -  2)  =  2  (2x  +  1)  +  aj 

36.  13  (5  -  x)  =  3  X  +  1 

37.  5  X  -  8  =  6  (x  -  3) 


44  THE  FIRST  USE  OF  ALGEBRA 

38.  3  (2  a;  -  5)  =  4  a;  +  1 

39.  25  -  4  a;  =  2  (x  -  4)  -  3 

40.  7  -  X  =  3  (4  -  x)  +  1 

41.  2  (a;  +  5)  =  5  (5  -  x)  +  a;  -  3 

42.  4  +  X  4-  4  (5  -  x)  =  3a;  +  2  (4  -  a;) 

43.  2x  +  5  =5(x-2)+2x-  10 

44.  1  +  X  +  3  (2  -  x)  =  7  (3  -  x) 

45.  13  (8  -  x)  +  2  (x  --  6)  =  2  X  +  1 

46.  9  (x  +  1)  +  5  =  5  X  -F  6  (x  -  1) 

47.  5  (x  -  6)  +  21  =  3  (x  +  3) 

48.  3  (x  -  4)  =  4  (x  -  3)  -  17 

49.  11  (x  +  3)  -  100  =  9  (x  +  5) 

50.  15(7x-3)  =8(13x  +  7) -20 

Sum  of  the  Unknown  Numbers  Given 

39.  Model  H.  —  A  merchant  has  grain  worth  9  cents 
per  peck,  and  other  grain  worth  13  cents  per  peck.  In  what 
proportion  must  he  mix  40  bushels  of  grain  so  that  the 
mixture  may  be  worth  40  cents  per  bushel? 

Let  X  =  the  number  of  bushels  at  36  cents  per  bushel. 

Then  40  —  x  =  the  number  of  bushels  at  52  cents  per  bushel. 
36  x  =  the  number  of  cents  value  of  one  grade. 
62  (40  —  x)  =  the  number  of  cents  value  of  the  other  grade. 
®    36  X  +  52  (40  -  x)  =  1600 

@     36  a;  +  2080  -  52  x  =  1600  ®  same  values 

®  2080  -  16  a;  =  1600  @  same  values 

0  2080  =  1600  +  16  x      (3)  +  lQx 

©  480  =  16  a;  0  -  1600 

0  30  =  a;  0  -M6 

0  10  =  40-a;  40-0 

Ans.    30  bu.  at  36ff  per  bushel;  10  bu.  at  52jf  per  bushel 
Check:  30  x  .36  =  10.80 

10  X  .52  =    5.20 
16.00, 


SUM  OF  THE  UNKNOWN  NUMBERS  GIVEN         45 

EXERCISE  14 

1.  The  total  rent  of  two  houses  is  $75  per  month.  One 
of  these  was  idle  for  three  months;  the  aggregate  rental 
received  from  both  houses  for  that  year  was  $801.  What 
was  the  rent  of  each  house? 

2.  Twenty  coins,  quarters  and  half-dollars,  amount  to 
$5.75.     Find  the  number  of  each  kind. 

3.  108  coins,  dimes  and  cents,  amount  to  $4.32.  Find  the 
number  of  each  kind. 

4.  Fourteen  coins,  nickels  and  quarters,  amount  to  $1.70. 
Find  the  number  of  each  kind. 

5.  100  coins,  quarters  and  dimes,  amount  to  $20.20. 
Find  the  number  of  each. 

6.  A  bridge  broke  down  under  a  pressure  of  28,500 
pounds,  caused  by  a  crowd  of  200  people.  If  the  average 
weight  of  a  man  is  150  lb.  and  that  of  a  woman  is  120  lb., 
how  many  men  were  there  in  this  crowd? 

7.  I  bought  30  pounds  of  sugar  of  two  grades  for  $2.28; 
the  better  grade  of  sugar  cost  10  cents  per  pound,  and  the 
poorer  7  cents.  How  many  pounds  of  each  grade  did  I 
buy? 

8.  I  bought  15  apples  for  59  cents;  red  ones  at  5  cents, 
green  ones  at  3  cents.     How  many  of  each  kind  did  I  buy? 

9.  A  grocer  was  offered  $15  for  50  pounds  of  tea;  he 
furnished  25-cent  tea  with  35-cent  tea  so  as  to  make  $1  on 
the  transaction.  How  many  pounds  of  each  grade  of  tea 
did  he  sell? 

10.  A  merchant  has  one  grade  of  grain  worth  11  cents 
per  peck,  and  another  grade  of  grain  worth  15  cents  per 
peck.  In  selling  20-bushel  lots,  how  many  bushels  of  each 
grade  must  he  furnish  in  order  that  each  lot  may  average 
48  cents  per  bushel? 


46  THE  FIRST  USE  OF  ALGEBRA 

11.  A  storekeeper  has  $197  in  bills  of  $2,  $5,  and  $10;  3 
times  as  many  twos  as  fives.  He  has  40  bills  in  all.  Find  the 
number  of  each. 

12.  The  complement  of  an  angle  added  to  five  times  the 
angle  is  equal  to  its  supplement.     How  large  is  the  angle? 

13.  Three  times  the  complement  of  an  angle  added  to  the 
supplement  of  the  angle  gives  twice  the  angle  itself.  How 
many  degrees  are  there  in  the  angle? 

14.  A  line  5  feet  long  is  divided  into  two  unequal  parts. 
Seven  lines  equal  to  the  shorter  part  and  2  lines  equal  to 
the  longer  part  would  cover  a  distance  of  18  ft.  9  in.  How 
many  inches  are  there  in  each  part  of  the  line? 

15.  A  rectangular  plot  of  land  fronting  on  a  street  is 
divided  into  6  house  lots  by  lines  running  back  from  the 
street.  Each  house  lot  has  the  same  frontage,  and  has  a 
perimeter  of  241  feet.  The  whole  plot  has  a  perimeter  of 
745  feet.     Find  the  dimensions  of  each  lot. 

16.  A  storekeeper  has  $4.04  in  dollars,  dimes,  and  cents; 
44  coins  in  all.  He  had  9  times  as  many  dimes  as  dollars. 
Find  the  number  of  coins  of  each  denomination. 

Circles 

40.  A  circle  is  a  figure  on  a  flat  surface  made  by  a  curved 
line  every  point  of  which  is  at  the  same  distance  from  a 
point  within  called  the  center.  The  curved  line  of  the 
circle  is  sometimes  called  the  circumference. 

Any  circle  may  be  supposed  to  be  marked  off  into  degrees 
like  a  protractor.  A  central  angle  will  intercept  a  portion 
of  the  circumference  containing  the  same  number  of  de- 
grees as  the  angle,  whatever  the  size  of  the  circle;  but  in  a 
large  circle  this  portion  of  the  circumference  will,  of  course, 
be  longer  than  in  a  small  circle. 


CIRCLES  47 

41.  The  distance  from  the  center  to  the  circumference 
of  a  circle  is  called  the  radius. 

Any  line  drawn  between  two  points  of  a  circumference 
and  passing  through  the  center  is  a  diameter. 

Any  portion  of  the  circumference  is  called  an  arc.  Two 
radii  that  make  an  angle  of  30°  intercept  an  arc  of  30°. 

In  a  circle  having  a  circumference  of  30  inches,  an  arc  of  30°  would 
be  2^  inches  long  (y\%  or  xV  of  the  circumference);  while  on  a  circle  of 
100  inches  circumference,  the  same  angle  would  intercept  an  arc  8| 
inches  long. 

42.  Two  arcs  are  complementary  when  their  sum  is  90°, 
just  as  in  the  case  of  angles.  Two  arcs  are  supplementary 
when  their  sum  is  180°. 

EXERCISE  15 

1.  If  each  degree  of  a  circle  is  3^  in.  long,  how  long  is  an 
arc  of  4^°  in  that  circle?  How  many  feet  are  there  in  the 
circumference  of  the  circle? 

2.  Two  arcs  of  a  certain  circle  differ  by  13°;  one  of  them 
is  6|  in.  longer  than  the  other.  What  is  the  length  of  an 
arc  of  90°  in  that  circle? 

3.  Two  arcs  of  a  certain  circle  are  complementary,  and 
each  is  11  ft.  3  in.  long.  How  long  is  a  degree  of  that  circle? 
How  long  is  the  entire  circumference? 

4.  Two  arcs  of  a  certain  circle  are  supplementary;  one 
is  7  ft.  long,  the  other  28  ft.  How  many  degrees  are  there 
in  each  arc?     How  long  is  the  circumference  of  that  circle? 

5.  One  base  angle  of  a  triangle  is  three  times  the  comple- 
ment of  the  vertical  angle,  and  the  other  base  angle  is  20° 
less  than  the  vertical  angle.     What  are  the  angles? 

6.  One  side  of  a  railroad  track,  laid  in  a  circular  arc  of 
33°,  requires  13.8  tons  of  rails,  some  of  them  weighing  40 
lb.  to  the  foot,  the  rest  48  lb.     Each  rail  is  30  ft.  long  and 


48 


THE  FIRST  USE  OF  ALGEBRA 


extends  over  li°.     How  many  rails  of  each  kind  are  needed 
for  this  side  of  the  track? 

7.  In  the  figure  formed  by  a  stripe  and  a  transversal, 
three  of  the  even  angles  have  a  sum  15°  greater  than  two  of 
the  odd  angles.     Find  the  angles  and  draw  the  diagram. 

8.  In  Fig.  1,  XY  is  parallel  to  CA,  the  angle  Q  is  40° 
greater  than  A,  and  is  twice  as  large  as  C.    Find  the  angle  B. 


Fig.  1. 


Fig.  2. 


9.  In  Fig.  2,  PQ  is  parallel  to  ZX,  A  =  125°,  and  X  =  Y. 
Find  the  angle  Z. 

10.  An  arc  of  30°  is  17  in.  long.  What  is  the  length  of 
the  circumference? 

11.  What  is  the  central  angle  of  an  arc  3  ft.  long  on  a 
circle  whose  circumference  is  45  ft.  long? 

12.  With  a  circumference  of  36  ft.,  what  is  the  supple- 
ment of  an  arc  of  17  ft.?  What  is  the  complement  of  an  arc 
of  7  ft.? 

13.  An  arc  of  42°  has  a  length  of  84  ft.  Find  the  length 
of  a  quadrant,  and  the  central  angle  of  an  arc  of  100  ft. 

14.  If  the  quadrant  of  a  circle  is  7  ft.  long,  how  long  is  an 
arc  of  3°? 

15.  If  an  arc  of  17°  is  2  ft.  10  in.  long,  how  long  is  the 
circumference? 


CIRCLES  49 

16.  If  an  arc  of  9°  is  f  in.  long,  how  long  is  its  comple- 
ment?   Its  supplement? 

17.  If  a  central  angle  of  72°  intercepts  an  arc  21  in.  long, 
how  long  is  a  quadrant  of  that  circle? 

18.  If  a  central  angle  of  22J°  intercepts  an  arc  8  in.  long, 
how  long  is  the  complement  of  that  arc? 

19.  On  a  circumference  of  25,000  miles,  how  long  is  an 
arcof9°?    Of  1.8°? 

20.  If  an  arc  of  3f °  is  1  ft.  long,  how  long  is  the  semi- 
circle? 

21.  If  an  arc  of  1|°  is  88  ft.  long,  how  long  is  a  quadrant? 

22.  If  a  central  angle  of  9°  intercepts  an  arc  of  ^V  i^-i 
how  long  is  the  whole  circumference? 

23.  On  the  outer  edge  of  a  protractor,  marked  to  half- 
degrees,  the  successive  marks  are  .0125  in.  apart.  How  long 
is  the  semicircumference? 

24.  A  railroad  track  is  laid  so  that  the  line  halfway  be- 
tween the  rails  is  a  circular  arc  of  20°,  each  degree  being 
30  ft.  long.  A  total  of  382  ties  are  used  on  this  curve,  the 
first  part  of  them  being  spaced  15  in.  on  centers,  the  rest 
3  ft.  How  many  degrees  of  the  arc  are  in  the  part  where 
the  ties  are  close-spaced? 


CHAPTER  II 
APPROXIMATE  COMPUTATION 

43.  Certain  methods  in  arithmetic  are  of  especial  value  in 
working  with  numbers  that  are  obtained  by  measurement. 
Such  numbers,  as  has  been  shown  in  Section  15,  depend  for 
their  accuracy  upon  the  precision  of  the  instruments  used 
and  the  skill  of  the  person  using  them.  Computations  based 
upon  such  numbers  may  involve  a  great  deal  of  tedious  and 
useless  labor.  The  methods  described  in  this  chapter  will 
prevent  that  waste. 

Multiplication 

44.  Model  A.  —  Multiply  .7854  by  31.07. 

We  begin  to  multiply  from  the  left  side  of  the 

7854  multiplier,  instead  of  from  the  right. 

3J  Qy  To  obtain  the  first  partial  product,  we  are  really 

Qo   ceo  multiplying  by  30,  not  by  3.     The  multiplicand  is 

■  rropj^  approximately  equal  to  .8.     Now  30  X  .8  would  be 

054978      10x2-4  or  24. 

Using  common  sense,  then,  instead  of  a  mechani- 

24.402378      ^^1  rule,  we  fix  the  decimal  point  after  23  in  the 

first  partial  product. 

The  other  partial  products  succeed  each  other  as  in  the  other  method 

of  multiplying,  except  that  each  successive  product  is  displaced  one 

column  to  the  right,  instead  of  to  the  left;  because  the  value  of  each 

place  in  the  multiplier  decreases  by  the  ratio  xVj  instead  of  increasing 

by  the  ratio  10. 

Thus  the  4  of  the  second  partial  product  is  displaced  one  column  to 
the  right  of  the  2  in  the  first  partial  product;  and  the  8  in  the  third 
partial  product  is  displaced  two  columns  to  the  right,  once  for  the  0  and 
once  for  the  7  in  the  multiplier. 

One  reason  why  this  left-hand  or  natural  order  of  partial 
products  is  preferable  is,  that  the  first  partial  product  so 

50 


MULTIPLICATION  51 

obtained  is  always  the  largest,  and  furnishes  a  sort  of  trial 
product,  to  which  the  other  partial  products  are  added  as 
successive  corrections. 

The  student  should  always  use  the  first  partial  product, 
therefore,  as  a  rough  check  upon  his  work,  and  especially 
should  fix  the'  decimal  'point  in  it  before  writing  the  second 
partial  product;  and  he  should  make  sure  that  the  number 
thus  obtained  as  a  trial  product  is  what  might  reasonably  be 
expected. 

45.  The  greatest  advantage  in  using  the  natural  order  of 
partial  products  is  in  the  saving  of  labor  by  discarding  un- 
necessary figures. 

In  Model  A,  for  instance,  at  least  the  last  four  figures  are 
doubtful.  If  by  more  careful  measurements  the  multiplier 
could  be  carried  one  place  farther,  we  might  find  it  to  be 
31.066  or  31.074,  and  this  would  change  the  last  four  columns. 
Also  the  decimal  .7854  might  have  any  value  between  .78536 
and  7.8544. 

The  effect  of  this  doubtfulness  at  the  end  of  each  factor  is 
illustrated  here  by  using  a  star  instead  of  the  possible  addi- 
tional figure. 

7854*  '^^^  ^^^^  ^^^^  there  is  a  star,  representing 

31.07*  ^^  unknown  figure,  in  the  fourth  decimal 

23.562*  column  shows  that  everything  beyond  the 

7854*         first  five  figures  in  the  product  is  value- 

54978*     less,  and  should  be  rejected.     It  cannot  be 

assumed  that  the  doubtful  figures  would 

•  ^"^^o i  o  increase  the  product,  for  if  the  measurement 
had  been  carried  one  place  farther,  the  value  might  have 
been  less;  .78536  instead  of  .7854,  or  31.066  instead  of 
31.07. 

Moreover,  as  the  quantities  are  expressed  in  but  four 
figures,  the  results  should  be  cut  to  four  figures,  that  is,  to 
24.40. 


52  APPROXIMATE  COMPUTATION 

46.  In  order  to  shorten  the  work  of  multiplication,  much 
of  this  rejection  can  be  done  in  each  partial  product.  We 
must,  however,  be  careful  to  avoid  errors  of  rejection.  To  do 
this,  we  add  1  to  the  last  figure  we  write  down,  whenever 
the  figure  rejected  amounts  to  more  than  half  of  that  1. 

47.  The  work  of  multiplication  as  thus  amended  is  now 
given  step  by  step: 

.7854 
One  partial  product :  31 .  07 

23.562 

All  figures  here  are  obtained  from  those  in  the  data;  there  may  be  a 
little  doubt  about  the  last  figure,  but  no  figures  appear  in  the  doubtful 
column. 

.7854 
Two  partial  products :  31 .  07 

23.562 

785 

The  last  figure  of  the  multiplicand  (4)  is  marked  off  to  indicate  that 
numbers  derived  from  that  figure  in  this  partial  product  would  appear 
in  the  doubtful  columns.  Note  that  in  this  case  the  figure  which  we 
reject  is  less  than  5.     Consequently  we  may  reject  it  without  adding 

1  to  the  figure  in  the  next  place  to  the  left. 

/// 
.7854 

31.07 


Three  partial  products :  23 .  562 

785 
65 


24.402  or  24.40 

The  next  two  figures  are  now  marked  off,  as  there  is  a  zero  in  the 
multiplier.  Also,  in  multiplying  we  say  7X9,  since  .7854  is  nearer  .79 
than  .78.    7  X  9  =  63;  hence,  carry  6.   Then  49  +  6  =  55. 

Since  one  figure  of  the  multiplier  is  zero,  we  get  only  three  partial 
products. 


MULTIPLICATION  53 

The  computation  is  now  complete;  it  is  fair  to  assume  that 
our  result  is  accurate  to  four  figures,  as  the  given  numbers 
are  expressed  to  four  figures. 

All  figures  of  the  multiplier  should  he  used  in  the  computation. 

The  nearest  value  of  the  rejected  figure  should  be  used  for 
//  / 

multiplication;  9  for  854;  5  for  54,  etc. 

The  successive  stages  of  the  work  in  this  example  are  separated  and 
repeated  here  for  the  purposes  of  study  only.  Of  course  no  one,  after 
finding  the  first  partial  product,  would  begin  over  again  and  put  down 
the  first  two  partial  products  and  so  on.  All  of  the  actual  work  is  shown 
in  the  last  stage  of  the  model  problem. 

Check  this  multiplication  by  using  round  numbers  (30  X  .8 
=  24.)  and  if  desirable  by  interchanging  multiplier  and  mul- 
tipHcand  as  follows: 

31.07 

.7854 

21.749 

2  486 

155 

12 

24.402  or  24.40 

48.  Model  B.  —  How  much  house  coal,  weighing  45.6  lb. 
to  the  cubic  foot,  can  be  stored  in  a  space  32  ft.  2  in.  long, 
21  ft.  5  in.  wide,  and  12  ft.  10  in.  high? 

In  reducing  these  measurements  to  feet  and  tenths  of  a  foot,  it  is 
convenient  to  make  a  table  of  equivalents,  as  follows: 

Inches:         123456789  10        11 

Tenths:        12        2*      34567        8*         8         9 

The  starred  numbers  occur  where  the  measurement  in  inches  is  just 
half  way  between  two  measurements  in  tenths.  Here  it  is  customary 
to  choose  the  even  number.  Since  there  are  just  as  many  even  num- 
bers as  odd  numbers,  we  shall  be  gaining  in  the  long  run  as  much  as  we 
are  losing.  This  principle  is  to  be  adopted  in  all  our  computation  where 
the  figure  rejected  is  exactly  5. 

The  working  out  of  this  problem  is  left  to  the  student. 


54  APPROXIMATE  COMPUTATION 

EXERCISE  16 

Multiply: 

1.  3.142  by  144.0  6.  301.7  by  180.6 

2.  92.5  by  3.14  7.  496.3  by  13.57 

3.  703.  by  21.8  8.  .0188  by  .00379 

4.  .785  by  1.21  by  288.  9.  1.192  by  .3489 

6.  47.38  by  71.17  lo.   1.97  by  32.8  by  .00523 

11.  What  is  the  weight  of  328  steel  beams,  each  17.81  ft. 
long,  weighing  178.3  lb.  per  linear  foot? 

12.  A  steel  beam  is  20.83  ft.  long  and  weighs  42.7  lb.  per 
foot.     Find  its  weight. 

13.  A  foot  of  wire  weighs  .109  lb.  How  much  will  a  mile 
of  this  wire  weigh? 

14.  A  meter  is  39.37  in.  How  many  square  inches  are 
there  in  the  surface  of  a  window  one  meter  square? 

15.  A  city  lot  measures  44.35  ft.  front  by  77.23  ft.  depth. 
Find  its  value  at  $1.63  per  square  foot. 

Division 

49.  In  long  division,  the  numbers  subtracted  are  partial 
products  obtained  in  multiplying  the  divisor  by  the  succes- 
sive figures  of  the  quotient,  beginning  at  the  left.  These 
successive  partial  products  may  be  curtailed,  as  in  multipli- 
cation, when  they  project  into  the  doubtful  columns. 

60.   Model  C.  —  Divide  63.3  by  27.9.  _2 

27  9)63  3 
One  partial  product :  *     cc'o 

27.9)63.3 

55  8 

7  5 

5_6 

1  9 


Two  partial  products: 


DIVISION  55 

Here  we  have  marked  off  the  9  in  the  divisor,  since,  when  multiplied 
by  2,  it  would  come  into  the  doubtful  column  to  the  right  of  the  last 
given  figure  of  the  dividend.  But  there  is  a.  2  to  carry  from  the  doubtful 
columns  (since  2x9  =  18,  which  is  nearer  20  than  10  in  value),  making 
the  second  partial  product  56  instead  of  54. 

./    2.26 
27.9)63.3 

Three  partial  products:  -y-^ 

5  6 


1  9 

1  7 


Two  figures  are  marked  off;  5  is  carried  from  the  doubtful  columns. 
(Why?) 

xr/  2.267 
Four  partial  products :  27 . 9)  63 . 3 

55  8 

Ans.    2.27  I  ^ 

5  6 

Three  figures  are  marked  off.    Only  what  is  ■'■  " 

carried  from  the  doubtful  columns  can  now  ap-  ^    ' 

pear  in  the  partial  product.     The  next  figure  in  2 

the  quotient  must  be  chosen  so  as  to  enable  us  2 

to  carry  2.      7  X  279  =  1953,  and  8  X  279  =  '      ' 

2232;  evidently  7  is  more  nearly  correct  than  8. 

Since  2.267  is  nearer  2.270  than  2.260  in  value,  we  must  increase  by 
1  the  last  of  the  three  figures  which  it  is  allowable  for  us  to  retain  in 
the  quotient.    The  answer,  then,  is  2.27. 

Check  this  division  by  the  use  of  round  numbers  (60  -i- 
30  =  2) .  For  a  more  thorough  check,  multiply  the  quotient 
by  the  divisor,  not  the  divisor  by  the  quotient. 

Multiplying  the  divisor  by  the  quotient  is  not  quite  so  good,  because 
the  partial  products  thus  obtained  will  be  the  same  as  the  subtrahends 
used  in  the  process  of  division. 


2.267 

27.9)63.3 

7  5 

1  9 

2 

0 

561  APPROXIMATE  COMPUTATION 

51.  In  the  so-called  Italian  method  of  division,  the  sub- 
trahends are  not  written  down;  the  computer  performs  the 
multiplication  and  the  subtraction  at  the  same  time,  as  fol- 
lows: 

Two  nines,  and  five  are  twenty-three;  two  sevens 
and  two  are  sixteen,  and  seven  are  twenty-three;  two 
twos  and  two  are  six,  and  zero  are  six;  and  so  on. 

The  "shop"  method  of  subtraction,  which  is  used 
here,  requires  the  computer  to  emphasize  the  number 
that  has  to  be  added  to  the  subtrahend  to  give  the 
minuend.    This  is  also  called  the  Italian  method  of  subtraction. 

Students  who  are  ambitious  enough  to  acquire  this  trick  will  find  it 
economical  when  there  is  much  dividing  to  do. 

EXERCISE  17 

1.  A  meter  is  defined  by  United  States  law  as  39.370  inches. 
A  carefully  surveyed  line  in  France  is  283.72  meters  long;  how 
many  feet  long  is  this  line? 

2.  An  arc  of  39.3°  is  2.73  feet  long.  How  long  is  one  de- 
gree on  this  arc? 

Divide : 

3.  308.5  by  .5632    7.  15.02  by  .1177  ii.  110.4  by  .3561 

4.  708.  by  3.14        8.  .350  by  7.83  12.  17.65  by  1.509 

5.  256.0  by  .7854    9.  .309  by  8.47  13.  .707  by  88.1 

6.  3.142  by  1.303  10.  77.07  by  3.043  14.  3.32  by  885. 

Perform  the  operations  indicated : 

5.  2.303  -^  359.6 

16.  3.835  X  100.2  -^  3.142 

17.  39.37  X  39.37  X  3.142 

18.  2.236  X  1.414  X  3.162 

19.  4.63  X  4.63  X  4.63  -f-  2.00 

20.  729.  X  .533  X  729.  X  .533 


SQUARE  ROOT 


57 


Square  Root 

52.  The  product  obtained  by  multiplying  a  number  by 
itself  is  called  the  square  of  that  number;  and  the  number 
which  is  multiplied  by  itself  is  called  the  square  root  of  the 
product. 

If  a  number  is  divided  by  its  square  root,  the  quotient  is 
the  same  as  the  divisor.  If  either  the  quotient  or  the  divisor 
is  smaller  than  the  square  root,  the  other  must  be  larger. 
The  square  root,  then,  must  be  somewhere  between  the 
quotient  and  the  divisor. 

53.  Model  D.  —  Find  the  square  root  of  55.90. 

Since  this  number  is  between  49  and  64,  its  square  root  wiU  be  be- 
tween 7  and  8;  let  us  guess  7.5  and  try  it. 


7.453 


7.500)55.90 

52  50 

3  40 

3  00 

40 

38 

2 

7.478 

7.476)55.90 

52  33 

3  57 

2  99 

58 

52 

6 

The  square  root  of  55.90  must  be  between  7.500 
and  7.453.  Let  us  try  a  number  halfway  between, 
say  7.476. 


To  four  figures,  then,  the  square  root  of  55.90  is 
7.477,  a  result  which  can  be  verified  by  multipUcation 
(or  by  division,  as  before). 


EXERCISE  18 
Find  the  square  roots  of 

1.  57.76  4.   1383.84        7.  3.142  lo.  32.1 

2.  94864.  6.   10.7584        8.   125.9  ii.  537.3 

3.  0.001444       6.  32.1489        9.  277.7  12.   .123 


58  APPROXIMATE  COMPUTATION 

Find,  to  four  figures,  the  square  roots  of 

13.  2  16.    6  19.    8  22.    250 

14.  3  17.   7  20.    12  23.  245 

15.  5  18.   10  21.  45  24.  252 

Compare  the  results  of  Exs.  13  and  19.  Also  compare  the 
results  of  Exs.  14  and  20.  Exs.  15  and  21.  Exs.  18  and  22. 
Exs.  15  and  23.     Exs.  17  and  24 

54.  It  sometimes  happens  that  the  data  of  a  problem  can 
be  stated  exactly  in  figures.  For  example,  if  there  are  just 
a  dozen  men,  whose  aggregate  weight  is  1954  pounds,  the 
number  of  men  is  exact,  though  the  total  weight  is  an  ap- 
proximate number.  In  the  same  way,  the  number  of  degrees 
in  the  three  angles  of  a  triangle  is  exactly  180.  Numbers 
that  are  thus  theoretically  exact  are  expressed  with  as  many 
figures  (zeros)  as  may  be  necessary  to  match  the  precision 
of  the  other  data. 

The  weight  of  the  men  might  vary  an  ounce  or  so  from  the  amount 
stated,  without  the  fact  being  detected  by  the  most  careful  weigher; 
consequently  this  number  is  properly  regarded  as  approximate.  The 
number  of  men,  of  course,  and  the  number  of  degrees  in  all  the  angles 
of  a  triangle,  are  numbers  that  are  not  obtained  by  measurement,  and 
they  cannot  admit  of  fractional  error. 

55.  It  sometimes  happens,  again,  that  of  several  data  in  a 
problem,  one  may  be  much  less  accurate  than  the  rest.  For 
example: 

On  a  carefully  surveyed  right  of  way,  3728.7  feet  in  length, 
railroad  tracks  are  laid  with  rails  weighing  25.3  pounds  per 
foot.     How  much  steel  is  required  for  these  rails? 

Here  if  we  multiply  25.3  by  3728.7,  all  the  columns  after  the  third 
will  be  doubtful.  And  if  we  multiply  3728.7  by  25.3,  all  partial  prod- 
ucts after  the  third  will  be  missing,  leaving  the  same  columns  in  doubt 
as  before. 


SQUARE  ROOT  59 


25.3**  3728.7 

3728.7  25,3** 


759**  74574 

177  18644 

6  1119 

2  ** 


94300. 


94300. 


56.  The  considerations  that  apply  in  addition  are  slightly 
different.  Thus,  if  three  weights  are  to  be  added  together, 
such  as  783.4  lb.,  2.874  lb.,  and  .005632  lb.,  the  three  num- 

783.4  bers  will  have  figures  in  nine  different  col- 

2.874  umns.      The  last  two  numbers  will  have 

•  005632    figures  in  columns  that  are  doubtful  on  ac- 

786.3  count  of  the  limited  precision  of  the  first 

number.     Consequently  the  last  weight  will  not  affect  the 

result  at  all,  and  the  second  weight  might  just  as  well  have 

been  reported  as  2.9  lb.     The  sum  of  the  three  is  786.3  lb. 

57.  This  principle,  then,  is  to  be  constantly  borne  in  mind: 
The  accuracy  of  a  result  is  no  greater  than  the  accuracy  of  its 

least  accurate  datum.  \ 

EXERCISE  19 

1.  A  line  738.5  feet  long  is  divided  into  two  parts  which 
have  the  ratio  1.137;  how  long  is  each  part? 

2.  Two  supplementary  angles  have  a  ratio  of  2.003.  How 
many  degrees  are  there  in  each  angle? 

3.  Two  supplementary  arcs  differ  by  3.10°,  and  their  sum 
is  303.3  feet  long.     How  many  feet  are  there  in  each  arc? 

4.  A  meter  is  39.370  inches  long;  express  that  length  in 
feet  to  the  same  number  of  figures. 

5.  A  mile  is  5280  feet  long.  How  many  feet  are  there  in 
1.29  miles? 

6.  The  bedplate  of  an  engine  lathe  is  3.73  feet  long.  How 
many  inches  is  that? 


^0  APPROXIMATE  COMPUTATION 

7.  A  gallon  is  3.785  liters,  and  a  liter  of  dry  air  weighs  1.293 
grams.    What  is  the  weight  in  grams  of  a  gallon  of  dry  air? 

8.  A  gallon  is  3.785  liters,  and  a  liter  of  dry  carbon  dioxide 
weighs  1.977  grams.  What  is  the  weight  in  grams  of  2.513 
gallons  of  carbon  dioxide? 

9.  One  acre  is  equivalent  to  0.4047  hectares.  How  many 
hectares  are  there  in  9.538  acres? 

10.  One  pound  is  equivalent  to  0.4536  kilograms.  How 
many  kilograms  are  there  in  263.2  pounds? 

11.  One  radian  is  equivalent  to  57.296°.  How  many  de- 
grees are  there  in  3.172  radians? 

12.  One  cubic  meter  is  equivalent  to  1.3079  cubic  yards. 
An  excavation  of  35.4  cubic  meters  is  paid  for  at  $2.23  per 
cubic  yard.     How  much  does  it  cost? 

13.  One  hectare  is  equivalent  to  2.471  acres.  How  much 
would  a  French  farm  of  11.3  hectares  cost  at  $72  per  acre? 

14.  The  two  base  angles  of  a  triangle  have  a  ratio  1.77,  and 
the  vertical  angle  is  43.4°  greater  than  the  difference  of  the 
base  angles.     Find  the  three  angles. 

15.  Two  of  the  angles  made  by  a  stripe  and  its  transversal 
have  a  ratio  1.7.  How  many  degrees  are  there  in  each  acute 
angle? 

16.  In  the  figure  made  by  a  stripe  and  its  transversal,  if 
each  obtuse  angle  were  3.83°  greater  than  each  acute  angle, 
how  many  degrees  would  there  be  in  each  of  the  two  interior 
angles  on  the  same  side  of  the  transversal? 

17.  The  angle  at  the  vertex  of  a  triangle  bears  to  the  angles 
at  the  base  the  ratios  1.33  and  2.17.  Find  the  three  angles 
of  the  triangle. 

18.  Thp  angles  at  the  base  of  a  triangle  have  the  ratio  .625, 
and  the  angle  at  the  vertex  is  4°  more  than  three  times  their 
difference.    Find  each  angle  of  the  triangle. 


SQUARE  ROOT  61 

19.  A  certain  railroad  track  consists  of  a  straight  track 
275  ft.  long  and  a  curve  on  which  each  degree  is  16.67  ft. 
long;  it  requires  800  ft.  of  rails.  How  many  degrees  are 
there  in  the  curve? 

The  curve  of  a  railroad  track  is  a  line  halfway  between  the  two  rails. 
The  outer  rails  make  an  arc  which  is  really  longer  than  that  of  the  inner 
rails,  but  the  total  length  is  computed  as  if  both  rails  were  laid  on  the 
center  line,  midway  between  the  actual  positions  of  the  rails. 

20.  In  the  basket  of  a  balloon  are  two  men,  weighing  as 
they  stand  160  lb.  and  210  lb.;  provisions  weighing  8  lb.; 
maps  weighing  11  oz.;  and  instruments  weighing  7.23  lb. 
What  is  the  total  load  in  the  basket? 

21.  A  miscellaneous  bill  of  goods  purchased  from  a  mail- 
order house  contained  a  washing  machine  weighing  125  lb. ; 
3  bolts  of  cloth  weighing  9  lb.  each;  3  spools  of  thread 
weighing  1  oz.  each;  2  stoves  weighing  360  lb.  and  158  lb.; 
and  a  packing  case  weighing  40  lb.  What  was  the  freight  at 
55  cents  per  hundred  pounds? 

22.  A  man's  real  estate  holdings  consist  of  two  farms  of 
230  acres  and  157  acres;  two  village  plots  of  3  acres  and  2 J 
acres;  a  ranch  of  1700  acres;  and  a  city  lot  50  by  85  ft. 
What  is  the  total  area  of  his  holdings? 

23.  A  large  corporation  owned  a  mill  employing  3500  men, 
another  mill  employing  2800  men,  a  machine  shop  employing 
125  men,  an  office  force  of  56  men,  and  an  experimental  plant 
of  12  technical  experts.  How  many  men  were  there  on  the 
pay-roll? 

24.  On  a  vacation  trip  of  five  days,  a  traveler  used  an 
automobile  the  first  day  and  covered  290  miles,  and  a  bicycle 
the  second  day  for  32  miles.  He  tramped  up  and  down  a 
mountain  the  third  day,  6.5  miles.  The  fourth  day  he  spent 
at  the  hotel  and  walked  only  150  yards.  The  fifth  day 
he  took  a  train  home  and  traveled  368  miles.  What  was  the 
total  distance  covered  on  his  vacation? 


63  APPROXIMATE  COMPUTATION 

58.   Model  E.  —  Solve  17.2  {x  +  3.77)  =  3.86  +  43.1  x, 

©      17.2  (a;  + 3.77)  =  3.86  + 43.1  x 
@         17.2  x  + 64.9  =  3.86 +  43.  la;    ©  same  values 
®  61.0  =  25.9a;  ©-17.2x-3.86 

©  2.35  =  x  ©-^25.9 

Check:        17.2(2.35  +  3.77)  =  3.86  +  43.1  (2.35) 
17.2(6.12)  =  3.86  +  101. 
105.  =105. 
Reject  no  figures  until  the  actual  work  of  computation  begins. 
In  checking  examples  solved  in  this  way,  the  values  found  for  the  two 
members  of  the  equation  may  not  be  exactly  equal,  but  ordinarily  they 
should  not  differ  by  more  than  one  per  cent  if  three  figures  are  used. 

Solve  the  following  equations : 

25.  57.3  X  +  2.87  =  283.6 

26.  2.13  (x  +  17.2)  =  109.7 

27.  8.7  {x  -  5.6)  =  27.8 

28.  1.14  {x  +  1.07)  =  7.23  +  X 

29.  21.3  {x  -  13.5)  =  513  +  a; 

30.  7.2  {x  -  3.6)  =  56  -  a; 

31.  .36  (x  +  .23)  =  7.6  -  2.4  a; 

32.  2.8  (x  -  3.8)  =  X  -  3.3 

33.  5.7  X  +  2.3  =  3.3  (8.7,  -  x) 

34.  6.4  {x  -  .77)  =  .56  (x  +  83) 

35.  4.38  (2.16  X  -  7.55)  =  28.9 

36.  5.88  X  -  3.11  =  2.97  (x  +  6.33) 

37.  3.34  (x  -  1.11)  =  X  +  80.9 

38.  X  -  29.7  =  2.34  (3.45  x  -  74.0) 

39.  .37  (x  -  .38)  =  .49  (x  +  .57) 

40.  4.39  X  =  32.4  (.127  x  +  .00525) 

41.  11.9  =32.7  (.115  +  .0438  x) 

42.  6.62  -  5.47  X  =  11.1  (x  +  10.3) 

43.  5.32  -  2.37  (x  -  5.89)  =  0 

44.  73.7  -  4.90  X  =  .145  (x  +  73.2) 


CHAPTER   III 

MEASUREMENT  FORMULAS 

59.  Beside  being  used  for  abbreviating  the  explanation  of 
problems  in  arithmetic,  algebra  is  used  also  for  abbreviating 
the  statement  of  arithmetical  rules. 

Rectangles 

60.  The  area-number  of  a  rectangle  is  the  product  of  the 
length-numbers  of  two  adjacent  sides. 

In  the  preceding  statement  three  numbers  are  mentioned; 
each  can  be  represented  by  a  letter  as  follows: 
S  =  the  area-number  of  the  rectangle. 
a  =  the  length-number  of  one  side. 
b  =  the  length-number  of  the  other  side. 

Then  the  rule  can  be  stated  by  the  formula: 
S  =  ah, 

61.  Suppose  we  find,  by  measuring,  that  a  rectangular 
building  lot  is  93.57  feet  long  and  64.25  feet  wide.  The  area 
will  be  the  number  of  square  feet,  and  decimal  subdivisions 
of  a  square  foot,  that  could  be  marked  off  on  the  surface  of 
the  lot.  We  know  that  this  area  is  found  by  multiplying  the 
two  length-numbers  together,  but  it  is  interesting  to  prove  it 
in  the  following  way: 

A  strip  exactly  a  foot  wide  along  the  short  side  of  the  rec- 
tangle could  have  a  square  foot  marked  off  on  it  for  every 
whole  foot  in  the  length  of  the  side  (64  of  those) ;  and  a  tenth 
of  a  square  foot  for  every  additional  tenth  of  a  foot  in  length 
(2  of  those) ;  and  a  hundredth  for  every  additional  hundredth 

63 


64  MEASUREMENT  FORMULAS 

(5  of  those).  A  tenth  of  a  square  foot  here  is  made  by  mark- 
ing one  square  foot  crosswise  into  ten  equal  strips,  1  ft.  long 
and  iV  f^-  wide.  A  hundredth  of  a  square  foot  is  made  by 
splitting  each  of  these  narrow  little  strips  into  ten  very  narrow 
strips,  each  measuring  1  ft.  by  y^^  ft. 

Notice  that  the  area-number  of  this  unit-wide  strip  is  the 
same  as  its  length-number. 

The  foot-wide  strip  along  the  shorter  side  of  the  lot,  then, 
has  an  area  of  64.25  square  feet;  and  one  of  these  strips  can 
be  marked  off  on  the  rectangular  lot  for  every  whole  foot  in 
the  length  (93  of  them). 

If  we  split  one  of  these  foot-wide  strips  into  ten  equal  strips, 
one  of  these  narrow  strips  can  be  marked  off  on  the  surface 
of  the  lot  for  every  tenth  of  a  foot  in  its  length;  and  a  hun- 
dredth of  a  strip,  in  the  same  way,  for  every  hundredth  of  a 
foot  in  its  length.  So  that  there  will  be  in  the  whole  lot  93.57 
strips,  each  having  an  area  of  64.25  square  feet. 

Since  the  same  reasoning  may  be  applied  to  any  rectangle, 
it  is  evident  that  the  area-number  of  a  rectangle  is  the  prod- 
uct of  the  length-numbers  of  two  adjacent  sides. 

EXERCISE  20 

1.  Find  the  area  of  a  rectangular  brass  plate  3.28  in.  wide 
and  5.37  in.  long. 

2.  A  rectangular  city  lot  has  40.27  ft.  frontage,  and  is 
inclosed  by  just  217  ft.  of  fencing.  What  is  the  value  of  the 
lot  at  $2.25  per  square  foot? 

3.  A  strip  of  land  running  back  from  a  city  street  the  whole 
depth  of  the  lot  is  in  dispute  between  adjoining  owners.  The 
width  of  the  strip  is  8.723  in.,  and  the  area  in  dispute  is  57.21 
sq.  ft.    How  deep  is  the  lot? 

4.  The  front  of  a  lot  is  38.0  ft.,  and  the  area-number  ex- 
ceeds the  other  length-number  by  2156.  Find  the  depth  of 
the  lot. 


RECTANGLES  65 

6.  The  perimeter  of  a  square  is  738.4  ft.  What  is  the 
area? 

62.  A  surveyor,  as  will  be  shown  hereafter,  can  determine 
the  ratios  between  different  lines  with- 
out actually  measuring  the  lines  them- 
selves. 

For  example,  in  the  accompanying 
figure,  the  angle  X  could  be  measured 
in  degrees,  and  then  by  referring  to 
tables  the  surveyor  could  find  the  ratio 
of  6  to  a. 

EXERCISE  21 

1.  The  ratio  of  two  sides  of  a  rectangle  is  1.14  and  the  long 
side  is  39.3";  what  is  the  area? 

On  drawings,  5  ft.  4  in.  is  commonly  written  5'  4". 

2.  The  ratio  of  two  sides  of  a  rectangle  is  3.08  and  the 
short  side  is  7.46".    What  is  the  area? 

3.  The  ratio  of  two  sides  of  a  building  lot  is  0.385,  and  the 
long  side  is  103.4  ft.     What  is  the  area? 

4.  The  ratio  of  two  sides  of  a  building  lot  is  .472,  and  the 
short  side  is  43.7  ft.    What  is  the  area? 

5.  The  ratio  of  two  sides  of  a  floor  is  1.34,  and  the  area  is 
301  sq.  ft.     What  is  the  perimeter? 

We  shall  have  in  this  example  an  equation  consisting  of  x^  =  some 
number.  These  equations  will  be  studied  later  in  our  work,  but  for  the 
present  it  is  quite  correct  for  the  student  to  find  the  square  root  of  the 
number  for  the  value  of  x. 

6.  The  ratio  of  two  sides  of  a  rectangle  is  2.07,  and  the 
perimeter  is  137  in.     What  is  the  area? 

7.  The  ratio  of  two  sides  of  a  rectangle  is  1.57,  and  the  two 
sides  differ  by  just  83  ft.    Find  the  area. 

8.  The  ratio  of  two  sides  of  a  rectangle  is  .785  and  the  area 
is  1025  sq.  in.    What  is  the  short  side? 


66 


MEASUREMENT  FORMULAS 


9.  The  ratio  of  two  sides  of  a  rectangle  is  .0243,  and  the 
perimeter  is  500  ft.     Find  the  area. 

10.  The  ratio  of  two  sides  of  a  rectangle  is  .335,  and  the 
two  sides  differ  by  43.7  in.    Find  the  area. 

Rhomboids 

63.  The  proof  given  in  Section  61  is  valid  for  the  areas  of 
rectangles,  because  the  unit  squares  and  their  subdivisions 
fit  exactly  at  the  corners  of  the  figure.  This  is  due  to 
the  fact  that  a  rectangle  is  formed  by  two  stripes  perpen- 
dicular to  each  other.  The  proof  does  not,  however,  ap- 
ply to  a  figure  formed  by  two  stripes  inclined  to  each  other. 
Such  a  figure  is  called  a  rhomboid. 


Rectangle 


64.  In  the  rhomboid,  represent  by  a  the  length-number  of 
each  of  the  two  sides  that  lie  along  one  of  the  stripes;  and  by 
pi  the  breadth  of  the  stripe. 


Draw  two  lines  straight  across  the  stripe  from  the  ends  of  one  of  the 
lines  called  a.  This  makes  three  spaces  within  the  stripe;  two  of  them, 
which  we  call  X  and  Y,  are  triangles,  and  the  other  is  an  irregular  figure 
that  we  represent  by  Q.  These  letters  are  understood  to  represent  the 
area-numbers  of  the  different  spaces. 


RHOMBOIDS 


67 


Q  +  X  isa,  rectangle,  of  which  the  sides  are  a  and  pi.    Consequently 

®  Q  +  X  =  api 

A  tracing  of  triangle  X  will  exactly  cover  the  triangle  Y.  Conse- 
quently       ©  X  =  Y 

®  Q  -{-  Y  =  api     @  subst.  in  ® 

But  Q  +  F  is  our  original  rhomboid;  so  we  have  a  convenient  ex- 
pression for  the  area: 

®  5  =  api. 

65.  To  state  this  briefly  in  words,  we  need  new  tech- 
nical terms :  A  figure  is  said  to  be  inscribed  in  a  stripe  when 
all  its  vertices  are  in  the  sides  of  that  stripe. 

When  a  figure  is  inscribed  in  a  stripe,  a  side  of  the  figure 
that  lies  in  one  side  of  the  stripe  is  called  a  base  of  the 
figure,  and  the  breadth  of  the  stripe  is  then  the  altitude  on 
that  base. 

We  have,  then,  the  rule: 

The  area-number  of  a  rhomboid  is  the  product  of  the  length- 
numbers  of  a  base  and  the  altitude  on  it. 

66.  Since  the 
sides  of  a  rectangle 
are  perpendicular, 
the  base  in  one 
stripe  is  the  alti- 
tude in  the  other; 
so  that  whichever 
side  we  take  as  a 
base,  we  multiply 
the  same  numbers  /' 
for  the  area.    But 

this  is  not  the  case  /^ 

with  the  rhomboid. 

Represent  now  by  b  the  length-number  of  one  of  the  bases  of  the 
rhomboid  in  the  other  stripe;  and  by  252  the  breadth  of  that  stripe. 

Draw  two  lines  straight  across  the  stripe  from  the  ends  of  b.  It 
happens  here  that  we  have  four  spaces  within  the  stripe,  instead  of  three 


68  MEASUREMENT  FORMULAS 

as  in  Section  64.    Two  of  them  are  triangles,  which  we  call  R  and  Q; 
and  the  other  two  we  call  X  and  Y. 

Q  +  Z  is  a  rectangle  of  which  the  sides  are  h  and  p2.    Consequently 

©         Q-\-X  =  hp2 
A  tracing  of  the  triangle  X  +  R  will  just  cover  the  triangle  Y  ■}-  R. 
Hence,         0         X  +  R  =  Y  +  R 

0  X  =  Y  ■     ®  -R 

0  Q  +  F  =  6p2      0  subst.  in  0 

But  Q  +  F  in  this  figure  is  our  original  rhomboid;  so  we  have  for  the 

area  here 

0  S  =  bp2 

67.  For  every  rhomboid  there  are  two  formulas  for  the 
area;  two  pairs  of  numbers  such  that  the  product  of  either 
pair  will  give  the  area  of  the  rhomboid. 

S  =  api  =  hpi 

68.  A  parallelogram  is  any  figure  formed  by  two  stripes 
crossing  each  other,  whether  the  stripes  are  perpendicular  or 
inclined.  Rectangles  and  rhomboids  are  both  classed  as 
parallelograms. 

EXERCISE  22 

69.  Model  A.  —  The  area  of  a  rhomboid  is  1.23  sq.  ft., 
and  the  lengths  of  the  two  altitudes  are  1.02  ft.  and  2.74  ft. 
What  are  the  bases? 

Since  the  figure  is  a  rhomboid,  we  shall  use  the  rhomboid  formula. 
©        S  =  api  ©  S  =  bp2  Formula  (Section  67) 

0        S  =  1.23  0  ;S  =  1.23         Given 

0       pi  =  1.02  0  p2  =  2.74        Given 

0  1. 23  =  1. 02a        0      1.23  =  2.746     0  and  0  subst.  in  © 
0  1.21=  a  0      .449  =  6  0 -M.02,  0 -5- 2.74 

Hence,  with  pi  =  1.02,  o  =  1.21.    With  p2  =  2.74,  6  =  .449. 

Find  a  and  h  for  the  rhomboids  having  the  following  data: 

1.  >S  =  28,  pi  =2,  p2  =  3.5. 

2.  *S  =  360,  pi  =  12,  p2  =  15. 

3.  S  =  2560,  pi  =  32,  p2  =  25.6. 


RHOMBOIDS 


69 


4.  ,S  =  32.7, 

5.  S  =  55.14, 
e.  S  =  .389, 


pi  =  8.59, 
pi  =  10.16, 
Pi  =  .107, 


P2  =  6.21. 
P2  =  12.38. 
P2  =  .839. 


70.   Model  B.  —  The  altitudes  of  a  rhomboid  differ  by 
7.23  ft;  the  bases  are  29.7  ft.  and  43.2  ft.    Find  the  altitudes. 
Let  pi  =  the  number  of  feet  in  the  smaller  altitude. 

Then  pi  +  7 .  23  =  the  number  of  feet  in  the  larger  altitude. 

Since  the  area  is  the  same  with  either  side  taken  as  the  base,  the 
shorter  altitude  must  be  the  one  on  the  longer  base. 


(J)  api  =  hpi 

@  a  =  43.2 

®  b  =  29.7 

®  43.2  pi  =  29.7  (pi  + 7.23) 

®  43.2  pi  =  29.7  pi  +  215 

(6)  13.5pi  =  215 

®  pi  =  15.9 

®       pi  +  7.23  =  23.1 
With  a  -  43.2,  pi  =  15.9;  with  b  =  29.7,  pz  =  23. 1. 
Check  by  areas. 


Formula 

Given 

Given 

@  and  ®  subst.  in  ® 

®  same  values 

®  -29.7pi 

®  -J-  13.5 

®  +  7.23 


Find  pu  p2,  and  S  for  rhomboids  as  follows: 
7.   a  =  60,      h  =  72,       difference  of  altitudes   =  10. 


8.  a  =  8.1,     h  =  5.4, 

9.  a  =  30.6,  h  =  28.3, 

10.  a  =  3.46,  h  =  5.19, 

11.  a  =  .588,  h  =  .961, 


=  1.2. 
=  1.47. 
=  1.17. 
=  .102. 


ft 


71.   Model  C.  —  Two  sides  of  a  rhomboid  differ  by  11.2 
;  the  altitudes  are  7.08  ft.  and  23.7  ft.     Find  the  perimeter. 

Let  a  =  6  +  11.2;  and  the  larger  altitude  must  be  taken  with  the 
smaller  side. 

®  api  =  bpi 

®  pi  =  7.08 

®  P2  =  23.7 

®       (6  +  11.2)7.08  =  23.7  6 
Solving,  b  =  4.78  and  6+11.2  =   16.0;  then  the  perimeter,  2a  + 
26  =  32.0  +  9.56  =  41.6. 


Formula 

Given 

Given 

®  and  ®  subst.  in  ® 


70  MEASUREMENT  FORMULAS 

Find  a,  h,  S,  and  perimeter  for  rhomboids  as  follows: 


12. 

Pi  =  36, 

P2  =  48, 

difference  of  sides 

=  20. 

13. 

Pi  =  4, 

P2  =  5, 

(I 

tC          11 

=  1.5. 

14. 

Vi  =  3.83, 

P2  =  4.27, 

(C 

11      l( 

=  1.53. 

15. 

Pi  =  11.2, 

P2  =  16.5, 

<( 

it        tc 

=  10.1. 

16. 

Pi  =  .49, 

P2  =  .73, 

t( 

ti       a 

=  .26. 

72.  Model  D.  —  The  longer  base  and  the  longer  altitude 
of  a  rhomboid  are  26.3  ft.  and  18.2  ft.,  respectively;  the  perim- 
eter is  79.2  ft.     Find  the  area. 

Since  the  perimeter  =  2a  +  2  6,  a  +  6  =  half  perimeter. 
Then  |  (79.2)  =  39.6  =  26.3  +  6,  from  which  h  =  13.3. 


©            S  =  hv2 

Formula 

®           P2  =  18.2 

Given  (longer  alt.  with  shorter  side) 

®            6  =  13.3 

Given 

0           >S  =  (13.3)  (18.2) 

0  and  0  subst.  in  0 

®            S  =  242. 

0  same  values 

To  find  the  other  altitude: 

(6)            S  =  avi 

Formula 

0         242  =  26. 3  pi 

0  subst.  in  0 

0      9.20  =  pi 

0  ^26.3 

Find  areas,  and  missing  sides  and  altitudes  of  the  following 
rhomboids : 

17.  Longer  base  =7.5,  longer  alt.  =5,       perimeter  =  27. 

18.  Shorter  base  =56,  shorter  alt.  =49,     perimeter  =  288. 

19.  Shorter  base  =  15.7,  shorter  alt.  =23.4,  perimeter  =66.2. 

20.  Longer  base  =38.2,  longer  alt.  =32.4,  perimeter  =  138. 

21.  Longer  base  =  .728,  longer  alt.  =.418,  perimeter  =  2.52. 

Triangles 

73.  As  the  following  figures  show,  a  triangle  can  be  in- 
scribed in  three  different  stripes.  There  will  be  three  bases, 
and,  corresponding  to  them,  three  altitudes. 


TRIANGLES 


71 


If  we  take  any  two  of  the  three  stripes  belonging  to  a  tri- 
angle (for  example,  the  two  having  bases  a  and  h),  we  find 
the  space  common  to  the  two  stripes  is  a  parallelogram. 


/ 


^v     h 


(y  s  ^     / 


y- 


This  parallelogram  is  composed  of  two  triangles,  and  we 
shall  find  that  a  tracing  of  one  of  them,  being  rotated  on  the 
figure  without  overturning,  can  be  made  to  cover  the  other 
exactly.  But  one  of  the  triangles  is  the  one  we  started  with; 
if  we  represent  its  area  by  S,  the  area  of  the  parallelogram 
will  be  2  S, 

Let  us  represent  the  altitudes  on  the  bases  a,  h,  and  c,  by  pi,  pi,  and 
Pi,  respectively.     Then 
©         2S  =  api 

©  S  =  ^api  '     ©  -^  2 

In  the  same  way  S  =  hbp2',  and  if  we  take  the  stripe  along  c  instead 
of  that  along  a,  it  is  evident  that  S  =  ^  cps. 


74.  We  have,  then, 
as  the  rule  for  the  area 
of  a  triangle : 

The  area-number  of 
a  triangle  is  half  the 
product  of  the  length- 
numbers  of  a  base  and 
the  altitude  on  it. 


S  =  I  api  =  I  bp2  =  I  cps 


72 


MEASUREMENT  FORMULAS 


75.  It  is  evident  that,  if  two  parallelograms  are  inscribed 
in  the  same  stripe  and  have  equal  bases  thereon,  they  will 
have  the  same  area. 

If  two  triangles  are  inscribed  in  the  same  stripe  and  have 
equal  bases  thereon,  the  two  triangles  are  equal. 

If  a  triangle  and  a  parallelogram  are  inscribed  in  the  same 
stripe  and  have  equal  bases  thereon,  the  triangle  will  be  equal 
to  half  the  parallelogram. 


Thus,  in  the  above  figure,  X  =  Y,  but  X 
figure  below,  Q  =  R  and  Q  =  i  P. 

The  sign  5^  means  "is  not  equal  to." 


Z;  and  in  the 


EXERCISE  23 

1.  A  triangle  and  a  rectangle  are  inscribed  in  the  same 
stripe,  and  have  equal  bases  thereon;  their  areas  differ  by 
783  sq.  ft.     What  is  the  area  of  each? 

76.  Model  E.  —  A  triangle  has  two  bases  that  differ  by 
5.55  ft.,  and  the  corresponding  altitudes  are  15.00  ft.  and 
10.59  ft.     Find  the  area. 

Let  a  =  the  number  of  feet  in  the  shorter  base. 

Then  a  +  5.55  =  the  number  of  feet  in  the  other  base. 
Also  the  longer  altitude  must  be  taken  with  the  shorter  base. 


TRIANGLES 

© 

S  =  hapi  =  hbp2  =  h  cPi 

Formula 

® 

api  =  bp2 

®  X2 

® 

Vi  =  15.00 

Given 

0 

V2  =  10.59 

Given 

® 

15.00  a  =  10.59  (a +  5.55) 

@  and  ®  subst.  in  @ 

® 

15.00a  =  10.59a +  58.78 

0  same  values 

® 

4.41a  =  58.78 

®  -  10.59a 

® 

a  =  13.33 

©-^4.41 

® 

a +  5.55  =  18.88 

(D  +  5.55 

73 


Then  the  two  bases  are  13.33  and  18.88;  the  area  =  ^^  (13.33)  (15.00) 
=  100  sq.  ft. 

Check  by  product,  |  (18.88)  (10.59)  =  100. 

Find  two  sides  and  the  area  of  the  following  triangles : 

2.  pi  =  15  in.,       p2  =  7.2  in.,     difference  of  a  and  6  =  13  in. 

3.  pi  =4.8  ft.,      p2=20ft., 

4.  pi  =4.7  in.,  p2  =  7.3in.,  " 
6.  pi=39.2  cm.,  p2=52.4  cm.,  " 
6.  pi  =  .0487  in.,  p2  =  .0392  in.,        " 

77.  Model  F.  —  The  base  of  a  triangle  bears  to  the 
altitude  thereon  a  ratio  of  1.53;  and  the  area  of  the  triangle 
is  24.5  sq.  ft.     Find  base  and  altitude. 

Let  pi  =  number  of  feet  in  altitude. 

Then  1.53  pi  =  number  of  feet  in  base. 


a 

IC 

h 

=  19  ft. 

a 

(( 

b 

=  3.2  in. 

a 

(( 

h 

=  14.8  cm. 

a 

(( 

h 

=  .0152  in. 

©  S  =  ^api 

®  5  =  24.5 

®  24.5  =  H1.53pi)(pi) 

©  24.5  =  .765(pi)2 

®  32.0  =  (pi)2 

®  5.66  =  pi 

©  8.66  =  1.53  pi 


Check  by  area  formula,  S 


Formula 

Given 

@  subst.  in  © 

®  same  values 

©  -^  .765 

Square  root  of  ® 

®  X  1.53 
Ans.    Altitude,  5.66  in.;  base,  8.66  in. 
^  api. 


74  MEASUREMENT  FORMULAS 

Find  base  and  altitude  for  the  following  triangles: 

7.  Area  =  24  sq.  in.,  ratio  of  base  to  altitude  =  1.5. 

8.  ''      =  140  sq.  ft., 

9.  "      =  24.0  sq.  cm., 

10.  "      =  8.09  sq.  in., 

11.  "     =  .260  sq.  ft., 

78.  Model  G.  —  Two  altitudes  of  a  triangle  are  23.7 
in.  and  15.3  in.,  and  the  corresponding  bases  differ  by  3.28 
inches.     Find  the  two  bases. 


C( 

li 

u 

11 

=  .7. 

il 

11 

It 

u 

=  .83. 

(( 

11 

(I 

cc 

=  1.29. 

C( 

il 

11 

11 

=  .873. 

Let        6i  =  number  of  inches  in  one  base; 

bi  +  3.28  =  number  of  inches  in  other  base. 

©        S  =  lapi  =  ^hpi  =  ^  cpz 

Formula 

(D                                 api  =  bp2 

©  X  2 

®                                 pi  =  15.3 

Given 

®                                  P2  =  23.7 

Given 

®         (&1  + 3.28)  15.3  =  23.7  6i 

®  and  ©  subst.  in  ® 

(D            15.3  6i  +  50.2  =  23.7  &i 

®  same  values 

0                             50.2  =  8. 46i 

®  -  15.3  6i 

(D                             5.97  =  6i 

©  ^8.4 

®                             9.25  =  &i  + 3.28 

®  +  3.28 

Ans.     Bases  are  5.97  : 

in.  and  9.25  in.,  respectively. 

Check  by  area  formula. 

Find  two  bases  for  each  of  the  following  triangles: 

12.  Altitudes,   15  in.  and   16.8  in.;  difference  of  corre- 
sponding bases,  3  in. 

13.  Altitudes,  5.3  ft.  and  4.7  ft.;  difference  of  correspond- 
ing bases,  .80  ft. 

14.  Altitudes,  .255  in.  and  .431  in.;  difference  of  corre- 
sponding bases,  .287  in. 

Find  two  altitudes  of  the  following  triangles: 

15.  Bases,  16  ft.  and  25  ft.;  difference  of  corresponding 
altitudes,  5.4  ft. 


TRAPEZOIDS  .        75 

16.  Bases,  .77  cm.  and  .61  cm.;  difference  of  corresponding 
altitudes,  .11  cm. 

17.  Bases,  32.7  ft.  and  19.4  ft.;  difference  of  corresponding 
altitudes,  4.11  ft. 

Trapezoids 

79.  A  trapezoid  has  only  one  pair  of  sides  parallel.  It  can 
be  inscribed,  therefore,  in  only  one  stripe;  and  the  bases  in 
that  stripe  are  of  different  lengths.     There  is  but  one  altitude. 


Represent  the  length-numbers  of  the  two  bases  by  6i  and  62,  and  that 
of  the  altitude  by  p. 

The  trapezoid  can  be  divided  into  two  triangles  by  drawing  a  line 
from  corner  to  corner  (a  diagonal). 

Represent  the  areas  of  the  two  triangles  by  X  and  Y.     They  have 
the  same  altitude  p,  and  bases,  61  and  62,  respectively.     Their  areas  are 
®  X  =  i  61P 

©  F  =  I  62P 

Hence,        ©        X  +  F  =  |  6ip  +  I  62P        ©  +  © 

©        X  +  F  =  §  p  (61  +  62)         ®  same  values 

Thus,  for  the  area  of  the  trapezoid,  we  have  the  rule: 
The  area-number  of  a  trapezoid  is  half  the  product  of  the  length- 
numbers  of  the  altitude  and  the  sum  of  the  bases. 

S   =   ip{b,  +  h2) 

EXERCISE  24 
80.   Model  H.  —  The  bases  of  a  trapezoid  are  2.386  in. 
and  2.834  in.,  and  the  altitude  is  3.862  in.     Find  the  area. 


76  MEASUREMENT  FORMULAS 

®  S  =  hp(J>i  +  b2)  Formula 

@  fei  =  2.386  Given 

®  62  =  2.834  Given 

0  p  =  3.862  Given 

0  S  =  i  (3.862)(2.386  +  2.834)  @,  0,  and  0  subst.  in  0 

0  ,S  =  (1 .  931)  (5 .  220)  ©  same  values 

0  5  =  10.08  0  same  values 

Ans.    The  area  is  10.08  sq.  in. 

Find  the  areas  of  the  following  trapezoids : 

1.  p  =32  ft.,        6i  =  48  ft.,  62  =  77  ft. 

2.  p  =  5  yds.,       61  =  1.2  yds.,.  62  =  4.8  yds. 

3.  p  =  6.27  m.,     61  =  5.83  m.,  62  =  9.21  m. 

4.  p  =  83.8  in.,     61  =  55.7  in.,  62  =  38.2  in. 
6.   p  =  .278  cm.,  61  =  .553  cm.,  62  =  .839  cm. 

81.  Model  I.  —  In  a  stripe  2.38  in.  wide,  a  trapezoid  is 
inscribed  whose  bases  have  a  ratio  834,  and  whose  area  is 
9.88  sq.  in.     Find  the  bases. 

Let  61  =  number  of  inches  in  one  base;  and  .834  61  =  number  of 
inches  in  other  base. 

0  S  =  hpihi  +  h)  Formula 

0  S  =  9.88  Given 

0  p  =  2.38  Given 

0      9.88  =  I  (2.38)  (61  +  .834  61)  0  and  0  subst.  in  0 
Solving,  &i  =  4.53,  and  .834  61  =  3.78. 

Check  by  area  formula. 

Find  the  bases  of  the  trapezoids  with  the  following  data: 

6.  Width  of  stripe  =  12  ft.,       ratio  of  bases  =  2, 

area  =  288  sq.  ft. 

7.  Width  of  stripe  =  1.8  in.,      ratio  of  bases  =  1.5, 

area  =  9  sq.  in. 

8.  Width  of  stripe  =  3.27  in.,    ratio  of  bases  =  3.29, 

area  =  77.8  sq.  in. 


TRAPEZOIDS  77 

9.  Width  of  stripe  =  75.6  ft.,    ratio  of  bases  =  1.82, 

area  =  93200  sq.  ft. 

10.  Width  of  stripe  =  .428  cm.,  ratio  of  bases  =  5.07, 

area  =^  .409  sq.  cm. 

82.  Model  J.  —  In  a  stripe  700  feet  wide,  a  trapezoid  is 
inscribed  whose  bases  differ  by  31.8  feet,  and  have  a  ratio  7.83. 
Find  the  area. 

Let  bi  =  number  of  feet  in  one  base;  7.83  6i  =  number  of  feet  in  other 


®  7.83  6i  -bi  =  31.8 

0  6.83  6i  =  31 . 8  ®  same  values 

(D  6:  =4.65  0-5-6.83 

0  7.83  6i  =  36.4  0X7.83 

0  S  =  h  P  Q>i  +  h)  Formula 

0  S  =  h  (700)(4.65  +  36.4)    0  and  0  subst.  in  0 

0  /S  =  350  X  41 . 1  0  same  values 

0  aS  =  14400.  0  same  values 

Ans.  The  area  is  14,400  sq.  ft. 

Find  the  areas  of  the  following  trapezoids: 

r  11.   Width  of  stripe,  30  in.;  sum  of  bases,  120  in.;  ratio 

of  bases,  2. 

12.  Width  of  stripe,  6  ft.;  difference  of  bases,  6  ft.;  ratio 
I       of  bases,  1.75. 

13.  Width  of  stripe,  5.3  in.;  difference  of  bases,  4.3  in.; 
^      ratio  of  bases,  2.1. 

[.  14.   Width  of  stripe,  5.32  cm.;  sum  of  bases,  54.0  cm.; 

I      ratio  of  bases,  1.28. 

15.   Width  of  stripe,  .294  ft.;  difference  of  bases,  .0611  ft.; 
I      ratio  of  bases,  .838. 


78 


MEASUREMENT  FORMULAS 


The  Number  tt 

83.  The  length  of  the  circumference  of  a  circle  can  always 
be  found  by  multiplying  its  diameter  by  a  certain  famous 
number,  whose  exact  value  is  represented  by  the  Greek  letter  tt 
(pronounced  pi). 

This  exact  value  cannot  be  expressed  in  ordinary  figures, 
any  more  than  V2  can;  but  it  can  be  expressed  approxi- 
mately as  3.1,  3.14,  3^,  3.142,  3.1416,  3.14159,  etc.,  according 
to  the  requirements  of  the  problem  in  which  it  is  to  be  used. 

84.  The  area  of  a  circle  can  always  be  found  by  multiply- 
ing the  area  of  the  square  whose  side  is  equal  to  the  radius  by 
this  same  number  tt. 


I  i 

If  c  =  the  length  of  the  circumference  of  the  circle,  r  =  radius, 
2  r  =  diameter,  S  =  area  of  the  circle,  and  r^  =  area  of  square 
whose  side  is  r,  then  we  have  two  formulas: 
c  =  2  irr 

The  two  theorems  represented  by  these  formulas  have  been  known 
for  many  centuries,  and  will  be  proved  in  next  year's  work. 

EXERCISE  26 

1.  The  radius  ofa  circle  is  21.3  ft.    Find  the  circumference 
and  the  area. 

2.  The  circumference  of  a  circle  is  283  in.    Find  the  radius. 

3.  The  area  of  a  circle  is  1.83  sq.  in.    Find  the  diameter. 


THE  NUMBER  ir  79 

4.  The  area  of  a  circle  is  288  sq.  ft.  Find  the  circumference. 

5.  What  is  the  length  of  an  arc  of  21.3°  on  a  circle  whose 
radius  is  11.9  ft.? 

6.  On  a  circle  whose  radius  is  39.37  in.,  how  many  degrees 
are  there  in  an  arc  20.38  in.  long? 

7.  What  is  the  radius  of  a  circle  in  which  a  quadrant  is  5.8 
in.  longer  than  the  radius? 

8.  The  radius  of  a  circle  is  .707  ft.  Find  the  circumference 
and  the  area. 

9.  An  arc  of  38°  is  16  in.  long.  What  is  the  length  of  the 
radius? 

10.  An  arc  of  13.7°  is  25.8  ft.  long.  What  is  the  length  of 
the  radius? 

11.  One  base  of  a  trapezoid  is  1  ft.  less  than  double  the 
other,  and  the  altitude  is  4  ft.  The  two  triangles  into  which 
the  diagonal  divides  the  trapezoid  differ  by  12  sq.  ft.  Find 
the  bases  of  the  trapezoid. 

12.  One  base  of  a  trapezoid  is  2.46  in.  less  than  3  times  the 
other,  and  the  altitude  is  2.16  in.  The  two  triangles  into 
which  the  diagonal  divides  the  trapezoid  differ  by  21.  7  sq.  in. 
Find  the  bases. 

13.  The  bases  of  a  trapezoid  differ  by  4  ft.,  and  the  altitude 
is  3.128  ft.  The  two  triangles  into  which  the  trapezoid  is 
divided  by  the  diagonal  have  a  ratio  1.5.  Find  the  bases 
and  the  area. 

14.  The  bases  of  a  trapezoid  differ  by  1.053  ft.,  and  the  alti- 
tude is  11.22  in.  The  diagonal  divides  the  trapezoid  into  two 
triangles  that  have  a  ratio  1.771.    Find  the  bases  and  the  area. 

15.  The  two  bases  of  a  rhomboid  are  24  ft.  and  15  ft.;  the 
altitudes  differ  by  3  ft.     Find  the  area. 

16.  The  two  bases  of  a  rhomboid  are  28.3  ft.  and  17.2  ft.; 
the  altitudes  differ  by  11.5  ft.     Find  the  area. 


80  MEASUREMENT  FORMULAS 

17.  The  two  altitudes  of  a  rhomboid  are  7.37  ft.  and  5.08 
ft.,  and  the  sides  differ  by  5.84  ft.  Find  the  sides  and  the 
area. 

18.  The  two  altitudes  of  a  rhomboid  are  .0394  in.  and  .0288 
in.,  and  the  sides  differ  by  .0113  in.  Find  the  sides  and  the 
area. 

19.  The  altitudes  of  a  triangle  are  8",  10",  and  12";  the 
longest  and  shortest  sides  differ  by  5".  Find  the  three  sides 
and  the  area. 

20.  The  altitudes  of  a  triangle  are  1.24",  1.89",  and  2.81"; 
the  longest  and  shortest  sides  differ  by  3.15".  Find  the  three 
sides  and  the  area. 

Implicit  Formulas 

85.  A  formula  which  does  not  express  directly  the  value 
of  the  letter  sought,  but  gives  an  equation  from  which  that 
value  can  be  determined,  is  called  an  implicit  formula. 

A  formula  which  gives  the  value  of  the  required  letter 
directly  upon  substitution  is  an  explicit  formula.  Thus, 
the  formula 

is  an  explicit  formula  for  *S,  and  an  implicit  formula  for  p,  6i, 
and  62. 

86.  Model  K.  —  In  the  formula  S  =  ^  p  (61  +  62),  find 
p  when  S  =  29.4,  61  =  2.36,  h  =  3.52. 

©  29.4  =  I  p  (2.36 +  3.52) 

@  29 . 4  =  ^  p  (5 .  88)  ®  same  values 

®  29.4  =  p  (2.94)  0  same  values 

0  10  =  2?  ®^2.94 

EXERCISE  26 

1.  In  the  formula  S  =  i  p  (61  +  62),  find  61  when  S  =  690, 
p  =  15,  62  =  55. 

2.  In  the  formula  S  =  irr^  find  r  when  S  =  600. 


IMPLICIT  FORMULAS  81 

3.  In  the  formula  ^  =  J  p  (61  +  62),  find  p  when  S  =  23.8, 
61  =  3.57,  62  =  4.83. 

4.  In  the  formula  c  =  2  ttt,  find  r  when  c  =  1000. 

6.   The  area  of  a  trapezoid  is  1.132  sq.  ft.;  one  of  the  bases 
is  7.321  in.  and  the  altitude  is  19.36  in.     Find  the  other  base. 
Ushig  the  values  given,  evaluate  the  following  formulas: 

6.  n(4:L-\-d)  d  =  5;    L  =  30;    n  =  255. 

7.  i^^  L  =  6;    a  =  10;    tt  =  3.1416. 

8.  Tr(R  +  r){R-r)     R  =  25;     r  =  20;    tt  =  3.14. 
2pP 

P  +  P 


9.   ^^  p  =  10;     P  =   12. 


Rr 

{R  -^  r)  {m  —  1)  ' 

-,  11.   2  dnm  d  =  17326;     n  =  1440;     m  =  12.6. 

12.  /  -  ^  X  ^  /  =  .372;     d  =   14.4;     /i  =  29. 

cR' 

13.  ^   Z;^,  c  =  3;    R'  =  100;    i2  =  21. 

14.  j4-  Q  =  110;    r  =  5;    X  =  3.1416. 
^=       15.    '— ^ a  =  10;    6  =  17;     c  =  21. 

16.  \{-^—  +  -^)      a  =  .002;    6  =  .015;    n  =  6. 

2\n  — a      n— 5/  ' 


17.  -(^  -  1  j  7^  =  8;    i2  =  .75;    t  =  10. 

18.  ^^  +  ^'^  +  ^'^       R  =  100;    r  =  120;      R'  =  1000. 

19.  j-^  a  =  15;    6  =  28;    c  =  41; 

K  =  126. 


82  MEASUREMENT  FORMULAS 

fi  (W'  —  w') 


TF'  =  1390. 
W{h-2ti  +  t) 


21-   wr/.-9/,4-/^     P  =  1244;     TT  =  2;    ^  =  14.5; 


^1  =  11.5;    t  =  10. 

22.  J(!r-g)-s(r-0-5(5-r) 

s  =  5;    S  =  1;    W  =500;    w;  =240; 
^  =  0;    T  =  44;    ^  =  6;    r  =  5.5. 

23-      1    .   ^^  c  =  .002;    ^  =  20;    H  =  60; 

/i  =  3;    d  =  13.5;    D  =  1.5. 

24.   ^(gly)  ^  =  243;     r  =  16;    ^  =  12; 

Q  =  96;    Tf  =  125. 

=».  2,,. „..(._ '_+..) 

F  =  1000;   w  =  25.6;     T  =  100; 
g  =  20;    <  =  4;    s  =  5. 


CHAPTER  IV 
ADDITIONAL  TYPES  OF  PROBLEMS 

87.  An  algebraic  expression  that  forms  one  entire  member 
of  an  equation  may  have  its  value  changed,  provided  the  ex- 
pression that  forms  the  other  member  of  the  equation  has 
its  value  changed  in  the  same  way. 

For  example,  if  any  two  numbers  are  equal,  we  can  in- 
crease them  both  by  5,  10,  or  17,  or  any  other  number;  or 
we  can  multiply  them  both  by  any  number,  or  in  any  other 
way  apply  the  axioms  of  Sections  36,  37  (page  41).  The 
results  will  be  different  from  the  original  numbers  in  value, 
but  the  equations  obtained  will  be  true,  because  the  new 
numbers  will  be  equal. 

In  short,  to  reduce  an  equation,  we  may  change  the  value 
of  the  two  equal  numbers,  if  only  we  keep  them  equal;  but 
in  reducing  an  expression  which  is  not  a  member  of  an  equa- 
tion, we  are  not  permitted  to  change  its  value. 

Fractional  Equations 

88.  When  equations  arise  that  contain  fractions,  it  is 
necessary  to  multiply  both  sides  of  the  equation  by  numbers 
which  will  change  the  fractions  to  whole  numbers,  as  in  the 
following  examples. 

89.  Model  A.  —  Solve  the  equation  3  =  2. 

®    1  =  2 

0    X  =  6  ©  X  3 

83 


84  ADDITIONAL  TYPES  OF  PROBLEMS 

90.   Model  B.  —  Solve  the  equation  ^  =  ?• 


^35 

©      2y-l 

©  X3 

©     10  2/  =  G 

©  X  5 

©         2/=. 9 

©  -MO 

Check: 

f  X  T%  =  f 

One  line  of  the  work  in  Model  B  could  have  been  saved  if  we  had 
multiplied  both  sides  of  ©  by  15  at  once,  instead  of  first  by  3  and  then 
by  5. 


T*          O*          T* 

91.   Model  C.  —  Solve  the  equation  o  ~  o  =  7  " 

1 
2* 

© 

X         X         X          1 

2       3  "  4       2 

© 

2x      X 

©  X  2 

© 

3x-2x=^-3 

©  X  3 

© 

.-H-. 

©  same  values 

© 

2x  =  3a;-6 

©  X  2 

© 

2  re  +  6  =  3x 

© +  6 

© 

6  =  x 

©  -2a; 

Here  if  we  had  multiplied  ©  by  12, 
the  denominators,  we  should  have  had 

the  least  common 

multiple  of 

© 

X        X        X         \ 

2       3  ~  4       2 

© 

6x  -  4a;  =  3a;  -  6 

©  X  12 

© 

2a;  =  3a;  -  6 

©  same  values 

© 

2a;  +  6  =  3a; 

©  +  6 

© 

6  =  a; 

©  -2a; 

Check:    f  - 
3- 

2  =  |-i 

FRACTIONAL  EQUATIONS  85 

EXERCISE  27 
Find  the  value  of  x  in  the  following  equations : 
X  .  X  ,   ^  .  7a;— 6  ,  x     Zx—b  ,        ^ 

x  ^  x  ^   \  ^  3a;  —  7      Zx      x  —  5 

^•2  +  3  +  2  =  ^-^        '°-  —r~-ii=^- 

^•§--7=6  +  ^  11.   ^-g=2a.-10 

X  ,  X      1      X  -\-2  xH-l,a;  +  3 

^•4  +  6-3=^T.  la.   ^-+^  =  x-2 

xaja;^  x  +  3,a;  +  5,^-, 

"•2-5=2-3  13.  -^  +  ^-+5=20; 

6.^-^+17=0  ,4.2^-^  =  2-. 

_2      ic_5a;  — 8  a;  +  5      a:  — 5_3x 

"^^        3  "^  6  ~      36  ^^'   ~S~  "^  ~2~  ~  T 

2a;— 1      1       -.a;  ,3a;,  5a;  — 2     _      ,    . 

8.-2 3  =  ^  +  6  16.  «  +  ^  +  -^-=2x  +  4 

17.  3x+^  =  l^  +  2a; 

x-l,3x-l,l      4x-3,x-7 

19.  3a;  +  5  +  ^— =  — ^ —  +  5x 

^      .^         3a;-  1   ,   15a;-  3  ,   „ 

20.  10x= — ~ 1 h7a;-l 

^     ,   ^      x  +  2     ^        5a; -2 

21.  3a;  +  5 jr—  =5x — 

,^       2 -3a;      9  -  15a;  ,   ^   ,  ^ 

22.  10a;=  — ^ \-5-\-7x 


86       ADDITIONAL  TYPES  OF  PROBLEMS 

92.  Model  D.  —  I  bought  a  certain  number  of  picture 
postcards  at  2  for  a  nickel,  and  the  same  number  at  3  for  a 
nickeL  I  sold  them  all  at  a  uniform  price  of  2  cents  apiece, 
and  by  doing  so,  lost  25  cents.     How  many  did  I  buy? 

Let  X  =  the  number  of  cards  of  each  kind  bought. 

Then      2  x  =  the  number  of  cards  of  both  kinds  sold. 

-^  =  the  number  of  cents  paid  for  cards  at  3  for  5  cents. 

5  X 

—  =  the  number  of  cents  paid  for  cards  at  2  for  5  cents. 


4  a;  =  the  number  of  cents  received  for  all. 

©        ^  +  ^  =  4x  +  25 

©    10  X  +  15  X  =  24  X  +  150 

©  X  6 

©                25  X  =  24  X  +  150 

@  same  values 

©                     X  =  150 

©  -  24x 

Ans. 

150  cards  at  each  price. 

150  at  2  for  5        $3.75      300  at  2  cents        $6.00 

150  at  3  for  5          2.50              Loss                $  .25 

$6.25 

Check: 


EXERCISE  28 

1.  A  merchant  invested  $3000  more  in  one  shop  than  in 
another.  In  the  first  he  lost  |  of  the  capital  invested,  and 
in  the  second  the  profits  were  40  per  cent  of  the  capital  in- 
vested. On  both  shops  he  made  a  net  gain  of  $2080.  How 
much  did  he  invest  in  each  shop? 

2.  Two  lines  have  a  ratio  7,  and  J  of  the  larger  line  ex- 
ceeds double  the  smaller  line  by  12  inches.  Find  the  lengths 
of  the  lines. 

3.  Two  sides  of  a  rectangle  have  a  ratio  f ,  and  the  perim- 
eter is  237.0  meters.     Find  the  area. 

4.  Two  angles  have  a  ratio  2,  and  the  smaller  angle  is 
three  times  the  complement  of  their  sum.     Find  the  angles. 

6.  Two  angles  have  a  ratio  3,  and  double  the  smaller 

angle  is  the  complement  of  }  of  the  larger.     Find  the  angles. 

6.   A,  B,  and  C  shared  a  sum  of  money  so  that  A  had  $30 


FRACTIONAL  EQUATIONS  87 

more  than  half,  B  had  $33  more  than  a  fourth,  and  C  had 
the  remainder,  which  was  $16.     Find  the  shares  of  A  and  B. 

7.  A  had  $107  and  B  had  $45.  B  gave  A  a  certain  sum 
and  then  had  only  J  as  much  as  A.  How  much  did  B  give 
A? 

8.  From  a  cask  of  wine,  ^\  had  leaked  out,  f  had  been 
drawn  out,  and  there  were  11  gallons  left.  How  much  wine 
was  there  at  first? 

9.  A  dealer  bought  pineapples  at  the  rate  of  $4  for  7 
dozen,  and  sold  them  at  the  rate  of  $2  for  3  dozen,  gaining 
$4.     How  many  pineapples  did  he  buy? 

10.  I  had  a  herd  of  sheep  and  sold  a  half  interest  in  the 
herd,  and  a  half  interest  in  one  sheep  besides.  Then  I 
found  that  I  owned  17  sheep  more  than  J  of  the  herd.  How 
many  sheep  did  I  own  at  first? 

'  11.  A  woman  had  a  basket  of  eggs,  and  sold  J  of  them, 
with  i  of  an  egg  besides;  what  she  had  left  were  three  more 
than  half  of  what  she  had  at  first.  How  many  eggs  had  she 
at  first? 

12.  One  angle  exceeds  half  of  another  by  10°,  and  ex- 
ceeds the  complement  of  their  sum  by  9°.  Find  the 
angles. 

13.  In  a  stripe  If  inches  wide,  two  triangles  are  inscribed; 
their  bases  differ  by  2.58  feet,  and  their  areas  have  a  ratio 
|.     Find  the  areas. 

14.  Two  men  started  together  on  a  long-distance  bicycle 
ride.  At  noon  one  of  them  had  gone  ^  of  the  distance, 
while  the  other  had  gone  f  of  it;  and  they  were  then  30 
miles  apart.     How  far  had  they  planned  to  ride? 

15.  A  and  B  started  eastward  from  Palmer.  At  noon  A 
had  gone  f  of  the  distance  to  Boston,  B  had  gone  ^  of  it,  and 
they  were  2  miles  apart.    How  far  from  Palmer  to  Boston? 


88  ADDITIONAL  TYPES  OF  PROBLEMS 

Subtracting  Fractions  and  Parentheses 
93.   Model  E.  —  In  solving  the  equation 

10  a; =  — \-1x-  1 

notice  carefully  that  when  we  multiply  by  14,  the  fraction 

15  a;  —  3 

= becomes  a  whole  number,  which  must  be  subtracted 

from  the  preceding  term  of  the  equation. 

140a;  -  (30a:  -  6)  =  21  x  -  7  +  98x  -  14. 

Here  we  have  30  x  —  6  to  subtract  from  140  x. 

If  we  had  to  take  30  a;  from  140  a;,  the  remainder  would 
be  110  a;;  now,  however,  since  we  are  taking  6  less  than  30  a;, 
our  remainder  will  be  6  more,  or  110  x  +  6. 

The  equation  is  therefore  solved  as  follows: 


Q 

10.      l^-^-^  =  ^--l  +  7.       1 

© 

140x  -  (30x  -  6)  =  21x  -7  +  98.  - 

14©  X  14 

® 

110x  +  6  =  119x  -  21 

@  same  values 

0 

llOx  +  27  =  119  a; 

® +  21 

0 

27  =  9  a; 

0-llOa; 

0 

3  =  a; 

®  -9 

j5  /p  _j_  3 
If  the  first  member  of  the  equation  were  10  x = i 

we  should  apply  similar  reasoning. 

When  we  multiply  by  14,  this  expression  reduces  to 
140x-  (30x  +  6). 

To  take  away  30  x  from  140  x  gives  a  remainder  of  110  x. 

But  we  are  to  take  away  6  more  than  30  x,  so  that  the 
remainder  will  be  110  x  —  6. 

Our  conclusions  may  be  expressed  in  the  general  state- 
ment: 


SUBTRACTING  FRACTIONS  AND  PARENTHESES     89 

94.  In  removing  a  parenthesis  with  the  minus  sign,  all  terms 
within  the  parenthesis  are  to  be  changed  in  sign. 

When  we  speak  of  "changing  the  signs,"  we  mean  chang- 
ing all  the  plus  signs  to  minus,  and  all  the  minus  signs  to 
plus. 

When  no  sign  is  expressed  before  a  term,  the  plus  sign  is 
understood. 

Terms  containing  the  same  radical  factor  and  differing  only 
by  a  numerical  coefficient,  such  as  2  V3  and  5  V3,  are  similar 
terms. 

EXERCISE  29 
Remove  the  parentheses  and  simplify: 

1.  5x  +  7- (2a:  +  3)  7.   9 x  -  V^  -  {VS  -  x) 

2.  3  a;  -  5  -  (a:  +  2)  a.   ^S  -  9  x  -  (x  -  V3) 

3.  17x  +  13  -  (3x  -  5)        9.  A -35-  (90  -  A) 

4.  11  X  -  3  -  (2x  -  1)         10.    198  -  (§  C  +  32) 

5.  7x-2-(3x-4)  11.    l-(x-l) 

6.  9a;  -  13  -  (4  -  2x)         12.   180  -  B  -  (90  -  B) 

13.  180  -{A+B)-  (90  -  A) 

14.  A  4  5  -  180  -  (A-  90) 

15.  180  -  (A  +  B)  -  (90  -A)  -  (90  -  C) 

16.  360  -  (A  +  5  +  C)  -  (180  -  A  -  B) 

17.  V2  -  \/3  -  (2  VS  -  V2) 

18.  VIO  -  3  V3  -  (3  VIO  -  5) 

19.  3  -  V2  -  (5  V2  -  V3) 

20.  a;  -  V2  -  (\/2  -  y)     23.   1.73  -  (5  -  a;) 

21.  3  V2  -  (1  -  \/2)  24.   75  -  (V2  -  11) 

22.  3  V2  -  (1  +  V2)  25.  56  +  3  V2  -  (13  -  V3) 


90  ADDITIONAL  TYPES  OF  PROBLEMS 

EXERCISE  30 

Find  the  value  of  x  in  each  of  the  following  equations: 
,     x-1      x+S      2 


3. 


5,_^7^8U 


10  2  '      5 

a.  x+-^  =  3 g- 

X      re  —  5      a;  —  8 

g;  —  4      X  —  5      a; +  5 

2  8      ~~^ 


8. 


6  8  4 


x  +  7      3a;  +  l_7x-ll 
.  ®"       15  25      ~       45 

10.  5  X  +  20  -  2  (a;  +  2)  =  40  -  4  (x  -  15) 

11.  10  (a^  +  1)  -  (3 a:  +  5)  =  30  -  lOx  +  2  (x  +  35) 

12.  30  (x  +  6)  -  10  (a;  +  3)  =  311  -  3  re 
^13.  2  re  +  3  (re  +  9)  =  200  -  4  (90  +  3  re) 

14.  5  (re  -  6)  +  17  =  110  -  2  (re  -  5) 

15.  13  (re  -  13)  +  10  =  200  -  2  (75  -  re) 

16.  200  -  7  (60  -  5  re)  =  3  (2  x  -  11)  -  100 
IV.  50  re  -  10  (re  +  10)  =  40  +  re  -  8  (2  -  re) 

18.  274  (2  X  -  10)  -  163  (8  +  re)  -  48  re  =  0 

19.  7re  -  103  (re  -  17)  +  17  (3rc  -  103)  +  135  =  0 


THE  CONSTRUCTION  OF  EQUATIONS 

20. 

|-|(4x-25)=i(5x-61) 

21. 

2x-7-^-^  =  17-3^;3^ 

22. 

^,      a;-M7      7a;- 2  ,  8a; -13 
^^           2       -       3        '         9 

23. 

4a;-7      _      3a;- 11      a;+17 
5                         4                 6 

24. 

'".""-(-^V)— '» 

25. 

llx+19      Qx-5      6a;+l 
6                  3        ~       6 

26. 

8X-3      4x-7^5^      ^3 

27. 

7a;- 11      6a; +  5      9a;-5 
4                 5               22 

28. 

2x  +  7      3a;  +  4              . 
3                5       -^      ^ 

29. 

a;  +  3      2a;- 13      x      ^ 
5               11       ~  3 

91 


The  Construction  of  Equation 

95.  In  constructing  an  equation  for  a  problem  stated  in 
words,  the  first  thing  to  do  is  to  make  a  careful  list  of  all 
the  numbers  that  are  referred  to  in  the  problem  and  are  not 
actually  given  in  figures. 

The  next  step  is  to  pick  out  one  of  these  numbers  to  repre- 
sent by  X,  and  from  that  as  a  starting  point  to  construct 
abbreviations  for  the  other  numbers  in  the  list. 

When  this  has  been  done,  it  will  be  found  either  that 
(1)  one  statement  in  the  problem  has  not  been  used  in  making 
the  abbreviations;  or  that  (2)  there  are  two  abbreviations 
for  the  same  number. 


92       ADDITIONAL  TYPES  OF  PROBLEMS 

In  the  first  case,  the  unused  statement  may  be  expressed 
as  an  equation.  In  the  second  case,  the  two  abbreviations 
for  the  same  number  will  form  an  equation. 

96.  For  the  purpose  of  illustrating  the  construction  of 
equations,  certain  types  of  problems  have  been  used  in 
schoolbooks  for  many  years.  They  are  interesting  as  a 
sort  of  puzzle,  but  they  have  real  advantages  also  for  the 
mathematical  student;  not  the  least  of  these  advantages 
being  the  practice  they  furnish  in  picking  out  separately  the 
material  facts  in  a  narrative  paragraph. 

In  Exercises  31-34  the  student  is  expected  to  construct 
the  equations,  but  not  necessarily  to  solve  them. 


The  Problem  of  the  Digits 

97.  The  problem  of  the  digits  is  based  on  our  notation 
for  numbers.  A  number  of  two  figures  (or  digits)  is  equal 
to  ten  times  the  first  digit,  plus  the  second.  Thus  the  num- 
ber 57  (read  it  "five-seven,"  as  in  a  telephone  call)  means 
(5  X  10)  +  7,  not  5  +  7,  or  5  X  7. 

In  the  same  way,  if  x  were  the  first  digit  and  5  the  second, 
the  number  would  be  10  x  +  5.  The  same  digits  inter- 
changed would  give  the  number  50  +  x. 

Interchanging  the  digits  of  a  number  may  be  brought  about 
by  adding  another  number;  for  example,  57  +  18  =  75. 

98.  Model  F.  —  In  a  certain  number  of  two  digits,  the 
first  digit  is  double  the  second;  and  if  27  is  subtracted  from 
the  number,  the  order  of  the  digits  is  reversed.  Find  the 
number. 

Let        X  =  the  units  figure. 

Then  2  re  =  the  tens  figure. 

The  value  of  the  number  is  10  (2  x)  +  x. 

The  value  of  the  number  with  digits  interchanged  is  10  (x)  +  2  x. 


THE  CONSTRUCTION  OF  EQUATIONS  93 


® 

20x  +  x-27  =  10x  +  2x 

® 

21  X  -  27  =  12  X 

©  same  values 

® 

21  X  =  12  X  +  27 

0  +  27 

® 

9x  =  27 

0  -  12x 

0 

X  =  3 

0-9 

Ans. 

63. 

Check: 

6  = 
36  = 

2X3 

63-27 

EXERCISE  31 

1.  In  a  certain  number  of  two  digits,  the  first  digit  is 
three  times  the  second;  and  if  36  is  subtracted  from  the 
number,  the  digits  are  interchanged.     Find  the  number. 

2.  In  a  certain  number  of  two  digits,  the  second  digit  is 
three  times  the  first;  adding  54  interchanges  the  digits. 
Find  the  number. 

3.  In  a  certain  number  of  two  digits,  the  first  digit  is 
four  times  the  second;  subtracting  54  interchanges  the 
digits.     Find  the  number. 

4.  In  a  certain  number  of  two  digits,  the  first  digit  is 
twice  the  second;  subtracting  36  interchanges  the  digits. 
Find  the  number. 

5.  The  sum  of  the  two  digits  of  a  certain  number  is  11, 
and  if  27  is  added  to  the  number,  the  digits  are  interchanged. 
Find  the  number. 

6.  The  first  digit  of  a  certain  number  less  than  100  is  1 
less  than  double  the  second  digit;  and  if  18  is  taken  from  the 
number,  the  digits  are  interchanged.     Find  the  number. 

7.  The  sum  of  the  two  digits  of  a  number  is  9;  and  if  45 
is  added  to  the  number,  the  digits  are  interchanged.  Find 
the  number. 

8.  The  sum  of  the  two  digits  of  a  number  is  11;  and  sub- 
tracting 45  interchanges  the  digits.    Find  the  number. 


94  ADDITIONAL  TYPES  OF  PROBLEMS 

9.  The  difference  of  the  two  digits  of  a  number  is  one  less 
than  the  unit's  figure;  and  if  18  is  taken  from  the  number, 
the  digits  are  interchanged.     Find  the  number. 

The  fact  that  subtraction  interchanges  the  digits  shows  which  digit 
is  the  less. 

10.  The  smaller  of  the  two  digits  of  a  number  is  5  less 
than  3  times  the  larger  digit;  and  if  9  is  added  to  the  number, 
the  digits  are  interchanged.     Find  the  number. 

11.  The  difference  of  the  two  digits  of  a  number  is  one 
less  than  3  times  the  smaller  digit;  and  if  45  is  added  to  the 
number,  the  digits  are  interchanged.     Find  the  number. 

12.  The  sum  of  the  digits  of  a  number  is  11  less  than  3 
times  the  larger  digit;  and  if  27  is  added  to  the  number,  the 
digits  are  interchanged.     Find  the  number. 

13.  In  a  certain  odd  number  which  is  a  multiple  of  5,  the 
first  of  the  three  digits  is  3  less  than  4  times  the  second 
digit;  and  if  16  times  the  sum  of  the  digits  is  subtracted  from 
the  number,  the  remainder  is  3.     Find  the  number. 

14.  In  a  number  between  300  and  400,  the  last  two  digits 
differ  by  4;  and  if  the  first  two  digits  are  interchanged,  the 
number  is  increased  by  90.     Find  the  number. 

15.  A  number  between  100  and  200  -has  the  middle  digit 
1  greater  than  the  last  digit;  if  the  number  is  divided  by  the 
sum  of  the  digits,  the  quotient  is  11.     Find  the  number. 

16.  With  two  digits  that  differ  by  two,  two  numbers  can 
be  formed  which  have  a  ratio  f .     Find  the  numbers. 

The  Problem  of  Two  Velocities 

99.  The  problem  of  two  velocities  is  based  upon  the  fact 
that  a  body  may  possess  two  motions  at  the  same  time. 

For  example,  a  ball  has  two  motions  if  it  is  rolled  along  a 
car  floor  while  the  train  is  in  steady  motion,  for  it  has  its 


THE  CONSTRUCTION  OF  EQUATIONS  95 

own  rolling  motion  in  addition  to  the  motion  of  the  train. 
The  ball  may  go  faster  than  the  train,  if  it  is  rolled  forward; 
or,  if  rolled  toward  the  rear  of  the  train,  it  may  still  be 
moving  forward,  in  the  direction  opposite  to  the  thrower's 
aim,  but  more  slowly  than  the  train  moves.  Thus  if  the 
train  is  running  44  feet  per  second  westward,  and  the  ball  is 
rolled  with  a  velocity  of  5  feet  per  second  towards  the  rear 
of  the  car,  the  ball  is  actually  moving  through  the  country 
westward,  39  feet  per  second;  if  it  is  rolled  forward,  its 
total  westward  speed  is  49  feet  per  second. 

The  problem  of  two  velocities  is  a  very  simple  illustration 
of  an  important  principle  of  physics  —  the  Composition  of 
Motion. 

100.  The  construction  of  equations  under  this  problem 
depends  upon  the  formula 

s  =  vt 

where  s  represents  the  number  of  miles  traveled  in  t  hours 
by  a  body  moving  at  the  rate  of  v  miles  per  hour. 

Before  beginning  the  following  exercise,  the  student  should 
invent  numerical  illustrations  of  this  formula.  For  example : 
If  I  walk  3  miles  an  hour  for  7J  hours,  I  travel  22|  miles. 
If  a  train  goes  400  miles  in  16  hours,  its  average  speed  is  25 
miles  per  hour.  If  a  steamboat  keeps  up  its  standard  speed 
of  18  miles  per  hour,  a  trip  of  192  miles  will  take  it  10  hours 
and  40  minutes. 

101.  The  model  on  page  96  will  serve  for  further  illustra- 
tion of  the  relation  between  velocity  and  distance  covered. 


96  ADDITIONAL  TYPES  OF  PROBLEMS 

Model  G.  —  The  fore  wheel  of  a  carriage  has  a  circum- 
ference of  8  feet,  and  the  rear  wheel  of  12  feet.  When  the 
fore  wheel  has  made  25  turns  more  than  the  rear  wheel,  how 
far  has  the  carriage  gone? 

Let  X  =  number  of  turns  by  the  rear  wheel. 

Then  a;  -j-  25  =  number  of  turns  by  fore  wheel. 

®  8  (x  +  25)  =  12  a; 

0  8  X  +  200  =  12  a;  0  same  values 

0  200  =  4a;  0-8a; 

0  50  =  X  0^4 

0  75  =  X  +  25  0  +  25 

Since  12  feet  is  the  circumference  of  the  wheel  which  made  50  turns, 
the  distance  is  12  X  50  =  600  feet.  Aws.     600  ft. 

Check:     12  X  50  =  8  X  75  =  600 


EXERCISE  32 

1.  A  man  can  walk  3  miles  an  hour,  or  ride  8  miles  an 
hour.  How  long  will  it  take  him  to  go  132  miles,  if  he  walks 
half  the  time  and  rides  half  the  time? 

2.  A  man  can  walk  4  miles  an  hour,  or  ride  7  miles  an 
hour.  How  long  will  it  take  him  to  go  66  miles,  if  he  rides 
half  the  time  and  walks  half  the  time? 

3.  A  man  can  walk  3.43  miles  an  hour  on  an  average,  and 
he  can  ride  a  bicycle  at  the  rate  of  11.1  miles  an  hour.  How 
long  will  it  take  him  to  go  40  miles,  if  he  walks  half  the  time 
and  rides  half  the  time? 

4.  A  street  car  travels  8.7  miles  an  hour,  and  a  cab  travels 
6.2  miles  an  hour.  A  man  goes  30  miles  with  time  equally 
divided  between  these  two  modes  of  traveling.  How  long 
does  it  take  him  to  make  this  trip? 

6.  The  fore  wheel  of  a  carriage  has  a  circumference  of  5 
feet,  and  the  hind  wheel  of  8  feet.    When  the  fore  wheel  has 


THE  CONSTRUCTION  OF  EQUATIONS  97 

made  20  turns  more  than  the  hind  wheel,  how  far  has  the 
carriage  gone? 

6.  The  fore  wheel  of  a  carriage  has  a  circumference  of  6 
feet,  and  the  hind  wheel  of  10  feet.  When  the  fore  wheel 
has  made  25  turns  more  than  the  hind  wheel,  how  far  has 
the  carriage  gone? 

7.  The  fore  wheel  of  a  carriage  makes  29  turns  more  than 
the  hind  wheel  in  going  a  certain  distance.  The  circumfer- 
ence of  the  fore  wheel  is  10.8  feet,  and  that  of  the  hind  wheel 
17.1  feet.     Find  the  distance. 

8.  Two  wheels  are  belted  together,  one  being  9.72  feet 
in  circumference,  and  the  other  14.3  feet  in  circumference. 
How  far  has  the  belt  traveled  when  one  wheel  has  made  10 
turns  more  than  the  other? 

9.  A  naphtha  launch,  which  averages  9  miles  an  hour 
when  in  running  order,  broke  down  when  running  out  to 
sea.  It  had  to  be  rowed  back  at  the  rate  of  |  of  a  mile 
an  hour,  and  reached  the  pier  just  32  hours  after  starting. 
How  far  out  to  sea  had  the  launch  gone  when  it  broke 
down? 

10.  A  motor  boat  went  downstream  at  the  rate  of  9.7 
miles  an  hour,  and  up  the  same  stream  at  5.9  miles  an  hour. 
Ten  hours  were  taken  for  the  round  trip.  How  far  did  the 
boat  go  up  and  down  the  stream? 

11.  A  toboggan  goes  downhill  at  the  rate  of  20  miles  an 
hour,  and  is  hauled  back  at  2|  miles  an  hour.  If  half  an 
hour  is  taken  for  the  trip  down  and  back,  how  long  is  the 
coast? 

12.  A  boy  has  an  hour  for  exercise;  how  far  may  he  ride 
with  his  father,  at  the  rate  of  10  miles  an  hour,  before  getting 
out  to  walk  back?     He  can  walk  3J  miles  an  hour. 


98  ADDITIONAL  TYPES  OF  PROBLEMS 

102.  Model  H.  —  A  man  who  can  row  5  miles  per 
hour  in  still  water  finds  that  it  takes  him  5  hours  to  row  up- 
stream to  a  point  from  which  he  can  return  in  4  hours. 
How  fast  does  the  current  flow? 

Let  X  =  the  number  of  miles  per  hour  the  current  flows. 

Then  5  +  x  =  the  number  of  miles  per  hour  the  man  can  row  down- 
stream. 

5  —  X  =  the  number  of  miles  per  hour  the  man  can  row  upstream. 

4  (5  +  x)  =  number  of  miles  in  4  hours  downstream. 

5  (5  —  x)  =  number  of  miles  in  5  hours  upstream. 

©        5  (5  -  a:)  =  4  (5  +  x) 

0         25  —  5a;  =  20  +  4a;  ©  same  values 

©  25  =  20  +  9a;  ®  +  5a; 

©  5  =  9  x  ©  -  20 

©  ^=x  © -9 

Ans.     River  flows  f  mile  an  hour. 
Check:  The  man  goes  upstream  (5  —  |)  miles  per  hour,  in  5  hours  22| 
miles;  downstream  (5  +  f)  miles  per  hour,  in  4  hours  22|  miles. 

EXERCISE  33 

1.  A  certain  river  flows  2  miles  per  hour.  A  fisherman 
finds  that  he  can  row  upstream  a  few  miles  in  6  hours,  but 
it  takes  him  only  3  hours  to  come  back.  How  fast  does  the 
fisherman  row  in  still  water? 

2.  A  launch  goes  downstream  for  an  hour,  and  then  takes 
3  hours  to  get  back.  If  the  stream  flows  4  miles  an  hour, 
how  fast  can  the  launch  go  in  still  water? 

3.  The  tide  flows  6  miles  an  hour  in  a  certain  channel.  A 
steamer  makes  the  passage  between  two  lighthouses  in  this 
channel  in  4  hours  against  the  full  tide.  With  the  tide, 
the  steamer  makes  the  same  distance  in  2  hours.  How  fast 
does  the  steamer  go  when  the  tide  is  not  running? 

4.  A  boy  can  row  4  miles  an  hour;  it  takes  him  4  hours  to 
row  downstream  and  back  on  an  errand,  but  only  a  quarter 
of  the  time  was  spent  going  down.     How  fast  is  the  current? 


THE  CONSTRUCTION  OF  EQUATIONS  99 

5.  It  takes  a  fishing  dory  two  hours  to  go  down  the  harbor, 
with  the  tide  in  its  favor;  coming  back,  with  the  tide  adverse, 
the  dory  took  four  hours.  If  the  tide  is  not  running,  the 
dory's  speed  is  6  miles  an  hour.     How  fast  does  the  tide  run? 

6.  A  man  who  can  row  5  miles  per  hour  finds  that  it  takes 
him  3  times  as  long  to  go  up  a  river  as  to  go  down.  Find 
the  speed  of  the  current. 

7.  On  a  river  which  runs  1|  miles  an  hour,  it  takes  me 
twice  as  long  to  row  upstream  as  down.  How  fast  can  I 
row  in  still  water? 

8.  On  a  river  which  runs  2  miles  per  hour,  a  launch  can 
go  only  §  as  fast  upstream  as  down.  How  fast  can  it  go  on 
a  still  lake? 

9.  A  man  can  row  4  miles  per  hour  in  still  water;  going 
upstream  he  goes  only  f  as  fast  as  he  comes  down.  Find 
the  rate  of  the  current. 

10.  The  speed  of  a  ferryboat  is  observed  going  upstream, 
and  again  going  downstream;  the  ratio  of  these  two  speeds  is 
.7,  and  their  difference  is  6  miles  an  hour.  What  is  the  speed 
of  the  current?     What  is  the  speed  of  the  boat  in  still  water? 

11.  A  man  who  can  row  6  miles  per  hour  in  still  water 
goes  1  mile  farther  in  3  hours  coming  downstream  than  he 
does  in  4  hours  going  up.     Find  the  speed  of  the  current. 

12.  A  man  can  row  3  miles  farther  in  2  hours  downstream 
than  he  can  in  4  hours  upstream.  The  stream  flows  2  miles 
per  hour.     How  fast  can  the  man  row  in  still  water? 

13.  A  man  wishes  to  row  down  the  river  to  the  steamboat 
landing  and  back  in  4  hours.  It  takes  him  2  hours  to  get 
down  there;  he  starts  back  immediately,  and  at  the  end  of 
the  4  hours  he  is  4  miles  short  of  his  starting-place.  The  man 
can  row  3  miles  per  hour  in  still  water.  How  fast  does  the 
river  flow?    How  far  off  is  the  steamboat  landing? 


J 

100  ADDITIONAL  TYPES  OF  PROBLEMS 

The  Problem  of  Ability  and  Time 

103.  The  problem  of  ability  and  time  depends  upon  the 
fact  that  the  greater  the  power  employed  for  a  task,  the  less 
time  is  required  for  it. 

If  I  can  do  a  certain  piece  of  work  in  3  days,  and  you  can 
do  the  same  work  in  2  days,  both  of  us  working  together 
ought  to  be  able  to  finish  it  in  less  time  than  either  of  us 
alone;  it  will  not  take  5  days  then,  but  less  than  2  days. 
The  central  facts  here  are  that  I  can  do  J  of  the  work  in  1 
day,  and  you  can  do  J  of  it  in  1  day;  so  that  both  of  us  are 
able  to  do  |  of  it  in  1  day.  Hence,  it  is  easy  to  see  that  the 
Work  can  be  done  in  1^  days  by  both  of  us  together. 

104.  Model  I.  —  A  can  do  in  4  days  a  piece  of  work 
for  which  B  would  take  12  days.  How  long  would  it  take 
them  to  do  the  work  together? 

Let  X  =  the  number  of  days  in  which  they  can  do  the  work  together. 


Then  in  1  day  both  can  do  -  of  the  work 


and 
and 


U     11      <(           A            "        '< 

1   u 

4 

((         (( 

:    1    u 
12 

ti         i( 

®      i  +  ^  =  l 

©  Xa; 

®     3x  +  x  =  12 
®            4  a;  =  12 
®               x  =  3 

4  +  12  =  1 

0  X  12 

d)  same  values 

®  -h4 

Ans.    Three  days  for  both. 

Check: 

EXERCISE  34 

1.  A  can  do  a  piece  of  work  in  15  days;  B  can  do  the 
same  work  in  18  days.  How  long  will  it  take  them  to  do  it 
together? 


THE  CONSTRUCTION  OF  EQUATIONS      101 

2.  A  tank  can  be  emptied  by  three  taps;  by  the  first  alone 
in  80  minutes,  by  the  second  alone  in  200  minutes,  and  by 
the  third  alone  in  5  hours.  How  long  will  it  take  to  empty 
the  tank  if  all  three  taps  are  open? 

3.  A  bathtub  is  filled  in  40  minutes  and  emptied  by  the 
wastepipe  in  one  hour.  How  long  will  it  take  to  fill  the  tub 
with  the  wastepipe  open? 

4.  A  cistern  can  be  filled  in  12  minutes  by  one  pipe  alone, 
or  in  8  minutes  if  the  second  pipe  is  also  turned  on.  How 
long  will  it  take  to  fill  the  cistern  with  the  second  pipe  alone? 

5.  A  can  do  in  2 J  hours  a  job  which  B  can  do  in  If  hours 
and  C  can  do  in  3i  hours.  How  long  will  it  take  all  of  them 
together  to  complete  the  work? 

6.  A  does  f  of  a  piece  of  work  in  10  days;  then  he  calls 
in  B  to  help  him,  and  they  finish  the  work  in  3  days.  How 
long  would  B  take  to  do  the  work  by  himself? 

7.  A  can  do  a  job  by  himself  in  6  days,  B  can  do  it  in  10 
days,  and  both  together  with  C  to  help  them  can  do  it  in  2  f 
days.     How  long  would  it  take  C  to  do  it  alone? 

8.  A  man  can  walk  the  length  of  a  street  in  9  minutes, 
and  a  boy  can  run  that  distance  in  6  minutes.  If  they  start 
at  opposite  ends  of  the  street,  how  long  will  it  be  before 
they  meet? 

9.  Two  brothers  wished,  to  get  a  certain  sum  of  money, 
and  agreed  that  each  would  earn  half  of  it.  One  boy  earned 
his  half  in  10  days,  the  other  in  15  days.  If,  however,  the 
boys  had  put  their  earnings  together  into  one  fund,  in  how 
many  days  would  they  have  obtained  the  sum  of  money, 
assuming  that  each  earned  at  the  same  rate  as  before? 

10.  A  merchant  can  walk  to  his  place  of  business  in  40 
minutes,  or  he  can  ride  to  it  in  10  minutes.  How  long  will 
it  take  him  if  he  walks  half  the  time  and  rides  half  the 
time? 


102  ADDITIONAL  TYPES  OF  PROBLEMS 

11.  Starting  at  noon,  I  can  get  to  town  on  a  wagon  at 
1:48  o'clock;  if  I  walked,  I  could  get  there  at  2: 15  o'clock. 
If  I  agree  to  walk  half  the  time,  at  what  time  must  I  get  off 
the  wagon? 

12.  A  train  runs  from  Portland  to  Boston  in  4|  hours,  and 
another  train  runs  from  Boston  to  Portland  in  4J  hours. 
If  the  two  trains  start  at  the  same  time,  how  long  will  it  be 
before  they  pass  each  other? 

13.  In  Ex.  12,  if  the  trains  leave  Boston  and  Portland  at 
7:45  A.M.  and  8:00  a.m.,  respectively,  at  what  time  do  they 


CHAPTER  V 
TRANSFORMATIONS 

105.  It  is  often  necessary  to  change  the  form  of  a  formula 
or  other  algebraic  expression,  or  to  perform  some  algebraic 
operation  upon  it.  In  order  to  be  able  to  do  such  things 
intelligently,  we  must  investigate  the  laws  and  the  methods 
of  addition,  subtraction,  multiplication,  and  division,  in 
algebra,  to  find  how  they  differ  from  the  similar  laws  and 
methods  of  ordinary  arithmetic,  if  at  all. 

Negative  Numbers 

106.  The  expression  a  —  h  is  the  algebraic  symbol  for  the 
result  of  subtracting  h  from  a.  But  if  h  is  greater  than  a, 
the  subtraction  is  impossible.  The  question  is,  in  that  case, 
what  does  a  —  h  represent? 

107.  To  take  a  particular  case:  suppose  I  send  to  my 
grocer  an  order  for  10  quarts  of  berries.  If  the  grocer  has  a 
quarts  in  stock  he  will  have  a  —  10  quarts  left  after  filling  my 
order. 

Let  us  suppose  that  his  original  stock  was  7  quarts.  He 
would  send  me  those,  and  would  still  have  an  unfilled  order 
for  3  quarts.  The  situation  would  not  be  the  same  as  if  he 
never  had  had  any  berries,  or  any  orders  for  berries;  for  he 
now  has  an  unfilled  order  for  3  quarts. 

To  put  these  circumstances  into  arithmetical  shape: 
It  is  required  to  subtract  10  from  7.     The  operation  is 
impossible.     But  7  can  be  taken  from  7,  leaving  0,  and  there 
is  still  3  to  be  taken  from  any  number  that  mxiy  hereafter  be 

103  " 


104  TRANSFORMATIONS 

added  to  the  expression;  just  as  the  grocer  is  expected  to  fill 
his  order  from  any  goods  that  may  hereafter  come  in. 

We  express  this  condition  by  writing  a  minus  sign  before  the  figure 
that  remains  to  be  subtracted: 
0     7  -  10  =  -3 

This  means  that  3  is  to  be  subtracted  frcnr  pny  number  that  may 
hereafter  be  added  to  the  expression.  If  we  add  5  to  the  expression  in 
0,  we  have 

©7-10  +  5 3  +  5,  or 

®  -3  +  5  =  2 

If  we  add  3  to  the  expression  in  ©,  we  have 
®     7  -  10  +  3  =  -3  +  3,  or 
(D  -3  +  3  =  0 

108.  The  expression  —3  differs  from  all  expressions  known 
to  arithmetic  in  this  particular,  namely,  that  you  can  get  zero 
by  adding  to  it. 

109.  When  two  numbers  added  together  give  zero,  one  is 
called  the  negative  of  the  other;  but  generally  by  a  negative 
number  we  mean  a  number  preceded  by  the  minus  sign. 

That  is,  3  is  the  negative  of  —3,  just  as  much  as  —3  is  the 
negative  of  3;  but  if  we  are  asked  which  of  the  two  is  nega- 
tive, we  say  —3. 

110.  Terms  preceded  by  minus  signs  are  called  minus 
or  negative  terms ;  and  terms  not  preceded  by  minus  signs 
are  called  plus  or  positive  terms. 

It  may  be  said  that  any  positive  term  is  the  negative  of  the 
corresponding  minus  term;  the  apparent  contradiction  is 
explained  when  we  remember  the  definition. 

Addition 

111.  When  several  expressions  are  to  be  added,  we  con- 
sider them  as  one  expression  and  unite  similar  terms. 

In  algebraic  addition,  it  is  not  generally  advisable  to  take  the  trouble 
to  arrange  similar  terms  under  each  other. 


ADDITION  105 

EXERCISE  35 

Add  the  following  expressions: 

1.  3  X  +  2  +  4  X  +  7;  3  X  -  12;  1  -  10  X 

2.  5x  +  7x-\-ldx  -  5x;  3  -  17  x-\-2  -  Sx;  X-  5; 
—  X 

3.  S-]-2x  +  S  +  9x-7-x-4:-3x;x  +  d-Sx; 

2  X  -\-  1;  —  x;  —  X 

4.  3  +  5x+13x  +  17-8a;  +  5-10x;  4x  +  2- 
3x  +  7;  34  -  X 

5.  171  X  -  243  +  318  X  -  411  X  -  111  +  150;  38  x  + 
117;  301  -  100  x;  102*  x  -  299 

6.  3(x-5);  5(x-5);  8  (5  -  x) 

7.  17  (2  X  -  3);  11  (4  -  3  x);  (x  -  1)  3 

-^  8.   7  a  +  341  +  2  a  -  100  -  200  a  -  200;   10  (a  -  3); 

3  (3  -  a) 

9.  3  V3  +  7;  2  -  V3;  7  V3  -  5;  1  -  V3 

10.  4^3-6,17  -2  ^y^;  -3-  ^3;  5  ^3  +  1 

11.  5  VIO  +  3;  10  +  2  VIO;  5  -  VlO;  -3  VIO  -  10 

12.  2  (^3  -  7);  13  (5  -  2  v'3);  7  (5  V3  -  30) 

13.  13  X  —  4:y-\-z;4y  —  10  X  —  5  z;4:Z  —  4:X  +  4:  y—4: 

14.  X  +  V2;  X  -  2;  X  +  3  -  4  V2;  3  -  2  X  -  3  V2 

15.  2  2/  +  X  -  V3;  3  y-5  x  +  2  V3;  3-5  V3-2  x-y 

16.  V3  -  ^3;  2  V3  +  3;  3  ^3  -  5  -  V3 

17.  5  -v/2-2+2  V5;  3  V5-2  V2;  17  V2+17-2  V5 

18.  4  V3  -  5  ^2;  -  3  +  4  ^2  -  3  V3;  7;  20  -  V2 

19.  X  -  V2;  X  -  2  +  3  \/2;  5  -  a^2 

20.  -  131  X  -  100 y  -  z;-  (100 z  -  100 x);  50 x+150 y 

112.   If  we  make  no  mistake  in  reducing  the  examples  of 
Exercise  35,  we  get  sums  that  are  correct,  whatever  the  values 


106  TRANSFORMATIONS 

of  the  letters.  2x  -{-  Sx  =  5x  whether  x  stands  for  1,  for 
1000,  or  for  .001. 

If,  now,  we  substitute  1  for  each  letter,  each  of  the  given 
expressions  will  reduce  to  a  number.  We  may  also  substitute 
1  in  the  same  way  in  our  answer  to  the  example.  This  result 
should  be  equal  to  the  sum  of  the  others.  This  process 
enables  us  to  check  the  correctness  of  our  work. 

Thus,  adding  300  a;  +  17  2/  +  2  and  7  a;.  +  10  i/  -  9, 

we  obtain  307  a;  +  27  2/  -  7. 

By  substituting  a;  =  l,2/  =  l,  we  get  319  +  8  =  327. 

This  result  goes  to  show  that  the  algebraic  addition  is 
correct. 

The  symbols  Vs  and  VlO  stand  for  exact  values,  just  as  letters 
would.  Consequently,  for  the  purpose  of  checking  our  work,  we  may 
substitute  1  for  Vs  or  any  other  root. 

113.  This  process  is  called  checking  by  coefficients,  be- 
cause in  substituting  1  for  a  symbol  used  as  a  factor  in  a 
term,  we  practically  leave  the  symbol  out  of  the  term  and 
find  the  sum  of  the  coefficients  in  each  expression. 

Checking  by  coefficients  is  not  an  absolute  check,  but  it 
enables  us  to  detect  some  possible  mistakes.  It  is  easy  to 
apply,  and  the  pupil  should  form  the  habit  of  using  it. 

Subtraction 

114.  A  minus  sign  preceding  a  parenthesis  indicates  that 
the  quantity  within  the  parenthesis  is  to  be  subtracted.  We 
have  already  seen  that  we  simplify  an  expression  containing 
such  a  parenthesis  by  removing  the  parenthesis  with  its  sign, 
changing  the  signs  of  the  terms  within  the  parenthesis,  and 
then  uniting  similar  terms  in  the  entire  expression. 

In  subtraction  the  minuend  is  the  number  from  which  the 
subtrahend  is  taken  to  leave  the  remainder. 

From  the  foregoing  it  will  be  seen  that  the  rule  for  sub- 
traction is : 


SUBTRACTION  107 

Change  all  the  signs  in  the  subtrahend  and  add  the  resulting 
terms  to  the  minuend. 

To  save  time  in  calculation,  the  signs  of  the  subtrahend  should  be 
changed  in  your  mind,  and  the  terms  should  be  united  without  writing 
the  expressions  again. 

EXERCISE  36 

Perform  the  following  subtractions: 

1.  From a;3  -  7  a;2  +  16  X  -  12  take  3^  -  dx" -\- 2x  -  4S. 

2.  From  a  —  h  take  a  —  h  —  c. 

3.  From  X  —  y  take  y. 

4.  From  h  take  —  s. 

5.  From  x3  +  9x2  +  2x-48  take  a^  -  4:X^  -  Sx -\- S. 

6.  Take  a^-5x2-2x  +  24  from  s^  +  2x^  +  4:X  +  3. 

7.  Subtract  24 -\- U  x  -  29  x'' +  Q  j^  from  2t'  +  3x^  - 
nx-  12. 

8.  Subtract  2  -  V3  from  6  +  5  V3. 

9.  Take  a  from  a  —  b. 

10.  From  9p  —  5g  +  4r  take  5q  —  2p-jr2r. 

11.  From  3  -  \/2  +  2  V3  take  1  +  3  \/2  -  V3. 

12.  From  V3  take' 5  -  2  V3. 

13.  Subtract  \/3  —  \/5  from  V5  —  \/3. 

14.  Subtract  x^  +  V3  from  3  \/3  -  x. 

15.  Take  3  V^^  -  a;  V2  from  2  Vx  +  S  V^- 

16.  2  a  -  2  6  +  3.C  -  d  -  (5  a  -  3  6  +  4  c  -  7  d) 

Note  that  the  sign  of  the  first  term  in  a  parenthesis  is  not  the  sign 
of  the  parenthesis.  Thus,  in  Ex.  16,  the  sign  of  the  parenthesis  is  — , 
but  the  sign  of  5  a  is  +.  When  the  parenthesis  with  its  —  sign  is 
removed,  the  sign  of  5  a  is  — . 

17.  x^  +  4:x^-2x^  +  7x-l-  (x^ +2x^-2  x^  +  Qx 
-1) 

18.  -  (a2  -  ax  +  x2)  +  3  a2  -  2  aa;  +  a;2 


108  TRANSFORMATIONS 

+  2X  +  8) 

20.  10  a^h-j-S  ah^-S  a^h^-V-  (5  a^b-G  aW-1  a^¥) 

21.  6  x'y-Z  xy^+7  y^-\-a^  -(8  x^y-3  xy^+9  y^+U  x^) 

22.  a;  +  1  +  (5  -  a:)  -  (3  +  14  x) 

23.  2  a  -  3  6  +  (3  a  -  2  6)  -  (2  a  -  3  6) 

24.  a  —  6  +  c  —  (a  +  ?)  —  c) 

26.  14  a  +  27  6  -  13  -  (7  a  -  110  &  -  17) 

26.  a;  -  V3  +  (\/3  -  \/2)  -  (3  V3  -  2  x) 

-  27.  A  -  90  +  (180  -SA)  -  (A  -  B) 

28.  X  V2  -  3  +  (3  a;  V2  +  5)  -  ( V2  -  \/3) 

29.  13  ^3  -  (5  -  10  ^y^)  -  (13  V3  -  \/S) 

30.  3.286  -  1.873  x  -  (11.93  -  28.45  x) 

31.  3  -  V2  -  (\/2  -  \/3)  +  (5  -  3  V3) 

32.  180  -  3/  -  (f  c  +  32)  +/  -  c 

33.  I  ci  +  32  -  (I  C2  +  32) 

34.  9  c  +  100  -  (5/  -  160)  +  (/  -  c) 
36.  lSO-\-  {A +B  +  C)  -  {C  -  A  -  B) 

Parentheses  within  Parentheses 

115.  Model  A.  —Simplify  3  x  -  2V3  -  [2  x  -  (SVS  + 
X  -  10)]. 

Sx  -2V^  -[2x  -  (5V^  +  x  -  10)1 
=  3  a:  -  2  V3  -  [2  X  -  5  V3  -  X  +  rO] 
=  3x-2V3-2x  +  5V3  4-a;-10 
=  2  X  +  3  v'3  -  10    Ans. 

116.  To  simplify  an  expression  in  which  parentheses  occur 
within  parentheses,  remove  them  one  at  a  time,  beginning 
with  the  inmost  parenthesis. 


The  vinculum  placed  over  an  expression,  as  2  x  +  1,  means  the  same 
the  parenthesis.    The  brace  { j  and  the  bracket  [  ]  are  also  used. 


MULTIPLICATION  109 

EXERCISE  37 

1.  X  -  (y  -  z)  +  X  {y  -  z)-\r  y  -  (z  +  x) 

2.  [k-is-t)]Ms-{t-k)]-lt-{k-s)]-{k+s+t) 

3.  2x-  {Qy+  \4.z-2x\)  -  (Qx-  \y  +  2zl) 

4.  -iQx-  {12y-4x)\  -  {Qy-  {4.x-7y)l 

5.  3  -  2  V3  -  [5  (2  -  V3)  -  V3] 

6.  10  +  5  VIO  +  [{x  -  10)  -  3(V10  -  5)] 

7.  V2  (3  -  V3)  -  2  S3  -  V3  (2  -  V2)  -  V6! 

8.  3.28  (1.403-7.86  x)-5.63  (2.87  x-[3.52x- 1.943]) 

9.  x{5-  V3)  -  {5-  (x  +  xV^)l 


''''l{l-^)-^\l-{l'l)\ 


Where  the  answers  to  the  following  equations  are  not  whole 
numbers,  express  them  in  decimal  fractions. 

11.  x  +  3  -  (2x  -  17)  =4      ' 

12.  2  X  -  7  -  (x  +  2)  =  2  X  -  (2  X  +  9) 

^13.   3x  +  2-(2x-5)  +  (9-7x)  =x+(2-5a;) 
~14.   13  -  (5  -  2  x)  -  (1  -  7  x)  =  25  -  (2  +  7  x)' 

15.  300  -  2  X  -  (115  -  3  x)  =  1000  -  (3  x  -  61) .    • 

16.  100  X  -  (39  +  8  x)  -  10  =  12  X  +  345 

17.  X  -  (3 X  -  2x+  1)  =  8  -  (3 x  +  7)   ^ 

18.  12  X  -  3  (x  -  2)  =  20  -  (5  +  9  -  7  x) 

19.  100  -  (2  -  15  X  -  23)  =  98  -  (20  X  -  24  -  5  x) 

20.  25 X  -  10  (1  +  x)  =  3  -  (x  -  5x  -  7)  +  8 x     . 

Multiplication 

117.  The  multiplier  and  the  multiplicand  are  called  the 
factors  of  the  product.  In  continued  multiplication  there 
are  more  than  two  factors. 

For  example,  3  X  5  =  15;  15  X  7  =  105;  105  X  11  =  1155. 

3  and  5  are  the  factors  of  15;  15  and  7,  or  3  and  5  and  7  are  the  fac- 
tors of  105;  3,  5,  7,  and  11  are  the  factors  of  1155. 


110  TRANSFORMATIONS 

The  factors  of  any  product  may  be  written  in  any  order. 
For  example,  3  X  2  X  4  =  2  X  3  X  4  and  5  a6  =  5  6a. 

118.  When  two  or  more  of  the  factors  of  a  number  are 
aUke,  it  is  sometimes  convenient  to  indicate  the  fact  as  fol- 
lows: 

2X3X3X3=2X33=54;2  aaahbhh  =  2  a%\ 

The  small  figure  denoting  the  number  of  equal  factors  is 
called  an  index  or  exponent. 

119.  The  product  of  equal  factors  is  called  a  power. 
One  of  the  equal  factors  of  a  power  is  called  a  root. 

For  instance,  27  is  the  third  power  of  3.  2  is  the  fifth  root  of  32.  64 
is  a  sixth  power,  2  being  the  root,  and  6  the  index;  64  is  also  a  third 
power,  4  being  the  root,  and  3  the  index;  and  64  is  a  second  power,  8 
being  the  root,  and  2  the  index. 

120.  The  second  power  is  generally  called  the  square, 
and  the  second  root  is  called  the  square  root.  7  is  the 
square  root  of  49  (7  =  V49). 

The  third  power  is  called  the  cube,  and  the  third  root  is 
called  the  cube  root.     5  is  the  cube  root  of  125  (5  =  vT^). 

The  expressions  a^,   ¥,   c*,   d^  are  read  respectively   "a 

square,"  "h  cube"  (or  third),  "c  fourth,"  "d  fifth."     p^  is 

read  "p  kth"  or  ''p  to  the  kth  power." 

Note  the  difference  in  meaning  between  a^,  a?,  a^,  and  02,  03,  ai.  The 
latter  symbols  are  read  "a  two,  a  three,  a  four." 

121.  The  degree  of  a  term  is  the  number  of  letters  that 
are  factors  of  it.  Thus  7  x^y^  is  a  term  of  the  fifth  degree. 
The  degree  of  an  expression  is  the  degree  of  its  highest 
term. 

122.  Model  B.  —  Multiply  5  a^y  by  3  xY^^.  ■ 

The  factors  of  the  product  are  5  xxxy  X  3  xxxxxyyyyzz.  These  can 
be  rearranged  so  as  to  bring  the  numerical  factors  together,  and  bring 
all  the  like  letters  together: 

5x3  xxxxxxxx  yyyyy  zz 
which  can  be  simplified  thus:      15  ci^i^z'^. 


MULTIPLICATION  111 

123.  It  is  evident  that  there  may  be  terms  that  use  the 
same  letters  as  factors,  such  as  5  x^y  and  5  xy^j  and  yet  are 
not  similar;  but  notice  that  these  terms  5  xxy  and  5  xyy  have 
different  sets  of  letters.  In  order  to  be  perfectly  clear,  let 
us  define  similar  terms  as  terms  which  have  the  same  sets  of 
letters. 

EXERCISE  38 
Multiply: 

1.  ZxX^y  6. 'Sa^b^  X4a362 
f2.  dxy  X7xy  7.   8  a^c  X  5  a^hc^ 

3.  3  ahc  X  ac  &.    12  ab'^d  X  15  a%c^ 

4.  a^  X  a2  9.   7  a^c^  X  4  a%(^ 

5.  a^^  X  a  lo.   a^  X  3  a^ 

Multiply  together  the  following  expressions: 

11.  x^^y;x^y^  14.    17  Vx;  2  y       17.    101  a;^;  2  a;ioi;  5  a: 

12.  rcioy^S'rrV^    15.   x^•2/^•X2/  is-    na^h;Sa¥;5a^b^ 

13.  2Sa^y;5x^y^   le.  Soi^;5y^;oi^;y^  19.   llp";7g^;2pg 

20.    7  a;7;  2  x^;  2  ^/^ 

124.  Checking  by  coefficients  is  of  little  use  in  multiplica- 
tion examples  such  as  these  in  the  preceding  exercise;  the' 
fact  that  in  each  example  the  degree  of  the  result  should  be 
the  sum  of  the  degrees  of  the  several  factors  may  be  used  as  a 
partial  check. 

125.  When  an  expression  is  separated  into  two  parts  by 
a  plus  or  a  minus  sign,  it  is  called  a  binomial. 

When  an  expression  is  separated  into  three  parts  by 
plus  or  minus  signs,  it  is  called  a  trinomiaL 

When  an  expression  is  separated  into  two  or  more  parts 
by  plus  or  minus  signs,  it  is  called  a  polynomial. 

When  an  expression  is  not  separated  into  parts  by  plus 
or  minus  signs,  it  is  called  a  monomiaL 


1 12  TRANSFORMATIONS 

The  expressions  multiplied  in  the  preceding  Exercise 
are  all  monomials;  the  multiphcation  of  binomials  and 
of  other  polynomials  depends  upon  the  multiplication  of 
monomials. 

126.  Model  C.  —  Multiply  5  a^  by  (3  a;  -  2  a). 

5  a2  (3  re  -  2  a)  =  5  a2  X  3  X  -  5  a2  X  2  a  =  15  a^x  -  10  a' 

EXERCISE  39 

Multiply: 

1.  7  X  (a;  -  1)  6.  2  (x  -  5) 

2.  3  a2  (a  -  x)  7.  ih  (4  k'^W  -  2  h'^j') 

3.  10  h^k  (2/1-5  /b2)  8.  101  h^k"-  (11  h''k'-21  h^k'') 

4.  11  p'q  (3  p^q  -  3  p^)  9.  38  hij  (23  hji  +  12  ijh) 

6.   7  X2/2  (2  X2/7  +  4  x^i/)        10.   10  ax  (100  6a;  +  1000  ex) 

Multiply  together  the  following  quantities: 

11.  3;  X  -  5;  2  13.    10x^;2y^;ixy^  +  i  xhj 

12.  3  a  (a  -  6);  2  a;;  3  ax         14.   3  a^h;  a^  -b^;  2ab 

15.  15  pq^;  9  qr^;  ^\  ph  —  ^V  V^ 

16.  7  ap2.  4  ct2pj  15  5gr2.  ^ly  ap  —  ^ly  ft^f 

The  Distributive  Law 

127.  The  pupil  has  noticed  that  where  a  polynomial  is 
multiplied  by  any  number,  each  term  is  multiplied  by  that 
number;  that  is,  3  (a;  —  5)  =  3  x  —  15. 

128.  This  principle  may  be  illustrated  again  as  follows: 
A  shopkeeper  sends  every  week  x  dollars  to  the  bank;  his 

messenger  uses  every  week  $1  for  necessary  expenses;  his 
wife  draws  out  y  dollars  for  her  personal  use;  his  son  remits 
every  week  z  dollars,  which  is  added  to  the  shopkeeper's 
account.  The  increase  of  the  shopkeeper's  account  each 
week,  then,  amounts  to  x  —  1  —  y  -\-  z.  In  5  weeks  the  in- 
crease would  be  5  (a;  —  1  —  2/  +  2!).    But  in  5  weeks  the 


MULTIPLICATION  113 

shopkeeper  would  send  5  x,  the  messenger  would  use  up  5, 
the  wife  would  draw  out  5  y,  and  the  son  would  remit  5  z. 
Therefore 

b  {x  —  1  —  y  -\-  z)  =  bx  —  ^  —  by  -\-  bz 

129.  This  Distributive  Law,  as  it  is  called,  is  one  of  the 
important  and  fundamental  principles  of  algebra.  It  applies 
also  to  minus  signs;  e.g.,  —  {x  —  z)  =  —x  +  z; 

and  —  (x  —  y-\-z  —  a  —  h-\-c)  =  — x-\-y  —  z-\-a  +  h  —  c. 

It  may  be  stated  as  follows: 

The  product  of  a  polynomial  by  a  single  term  is  found  hy 
multiplying  each  term  of  the  polynomial  successively. 

The  negative  of  a  polynomial  is  found  by  taking  the  negative 
of  each  term  successively. 

Distributive  Factoring 

130.  A  glance  will  show  when  a  monomial  is  a  factor  of 
every  term  of  an  expression;  such  an  expression  can  there- 
fore readily  be  separated  into  factors.     Thus  in  the  expression 

5  a2  -  15  a6  +  20  a3  -  5  a 

5  a  is  a  factor  of  every  term,  and  the  whole  expression  is  the 
product  of 

5  a  (a  -  3  6  +  4  a2  -  1) 


EXERCISE  40 

Find  the  factors  of  the  following 
you  cannot: 

1.   ax  -{•  ay           5.   a^  —  ab 

expressions,  or  tell  why 
9.   13  x  + 91 2/2 

2.  ax  —  ay           e.  x^  +  x^^y           lo.   x'^y  —  xy^ 

3.  5x+10i/       7.   a;  +  rr2              n.   7x+49x2+343x3 

4.  3  a;  -  15  ?/     ^.   5  x  +  25  a^       12.   3  a:+6  xy-\-2  yz 

13.  13  x3  _^  65  3.2^  4.  117  3-2^2 

14.  343  7?y  +  98  xV  +  28  xf  +  8  2/^ 

114  TRANSFORMATIONS 

15.  I5x^  +  9xy  +  25  y"- 

16.  38  x3  +  57  x^y  +  19  xy"^ 

17.  x^o  +  10  x^y  +  45  a;V  +  120  xY  +  210  xY 

18.  45  a^^i/s  -  120  xy  +  210  xV  -  270  x^ 

19.   71^  —  n  +  n^  —  n^  20.   rs  —  2  r^s  +  rsP'  +  r^s^ 

EXERCISE  41 

Substitute : 

1.  y  =  X  -\-  2  in  the  expression  3  x  +  4  ?/  —  25 

2.  2  =  2  ?/  —  3  in  the  expression  3  x  +  3  2;  —  21 

3.  z  =  S  —  y  in  the  expression  z  -\-  10  —  5y 

4.  ^=2x  +  3in  the  expression  x^  +  2  xy  +  ^/^ 

5.  X  =  y  —  1  in  the  expression  3  x  +  2  x?/  +  2  x^ 

Multiplication  of  Polynomials 

131.  In  multiplying  a  polynomial  by  a  polynomial,  the 
multiplicand,  considered  as  one  quantity,  is  multiplied  by 
each  term  of  the  multiplier;  then  these  partial  products  are 
added.     The  work  is  arranged  as  follows: 

Model  D.  —  Multiply  2a-75by5aH-3&. 

2a  -7h 
5a  +  3b 
10  a2  -  35  ab 
+  6  ab  -  21  b^ 
10  a2  -  29  a&  -  21  62 

132   Model  E.  —  Multiply  2x-5by3x-4. 

2x  -5 
3  a;  -  4 
6  x2  -  15  x 

-Sx  +20 


6  a;2  -  23  x  +  20 


The  second  term  of  the  multiplier  gives  a  product  which  is  to  be  sub- 
tracted, and  we  change  the  signs  of  the  terms  as  we  write  them  down; 
that  is,  for  -4  (2  X  -  5)  we  write  -8  x  +  20. 


MULTIPLICATION  115 

A  very  good  check  for  multiplication  is  to  multiply  the 
multiplier  by  the  multiplicand;  this  gives  a  different  set  of 
partial  products. 

133.  Model  E  illustrates  all  the  combinations  of  signs 
that  are  possible  in  algebraic  multiplication.  It  is  worth 
while  to  notice,  therefore,  that  when  the  multiplicand-term 
and  the  multiplier-term  have. like  signs,  the  product-term  is 
plus;  when  the  multiplicand-term  and  the  multiplier-term 
have  unlike  signs,  the  product-term  is  minus. 

These  rules, are  nothing  more  than  our  old  ways  of  dealing 
with  positive  and  negative  parentheses;  they  are  a  more 
convenient  way  of  saying  that  when  the  multiplier-term  is 
plus  we  do  not  change  the  signs  of  the  multiplicand,  and  when 
it  is  minus  we  change  them  all. 

Both  rules  are  generally  condensed  into  the  following: 

In  multiplication,  like  signs  give  plus,  and  unlike  dgns 
minus. 

The  same  rule  holds  for  division,  as  will  hereafter  be 
shown. 

134.  For  further  illustration  of  this  important  rule,  we  may 
represent  the  multiplication  of  {x  —  3)  (x  —  2)  by  a  diagram. 

Here  x^  is  represented  by  the  entire  square; 
(x  —  3)(x  —  2)  by  the  rectangle  X;  2  x  by  the   -j- 
rectangle  A  +  Q;  3  x  by  the  rectangle  B  +  Q;     \ 
and  6  by  the  rectangle  Q.  i 

To  find  the  rectangle  X,  which  represents  T 
{x  -  3)  (x  -  2),  from  x^  we  subtract  (A  +  Q),  T 
which  represents  2  x,  and  then  (B  +  Q),  which  f 
represents  3  x.  J^ 

But  we  have  thus  had  to  subtract  Q  twice; 
and  to  do  this,  we  must  have  added  Q  after  one 
of  the  subtractions. 

We  may  express  the  work  thus: 

(x-'3)(x-2)=x^-(A+Q)+Q-iB  +  Q) 
a;2-2x  +  6-3x 


X 


116  TRANSFORMATIONS 


EXERCISE  42 

Multiply: 

1. 

(x  +  3)(x  +  2) 

9.    (11  -  5  a)  (5  -  2  a) 

2. 

(x  +  3)  (x  -  2) 

-^10.    (101  -  17  x)  (11  +  3x) 

3. 

(x  +  5)  (x  -  3) 

11.    (x  +  1.32)  (x  -  5) 

4. 

(x  -  5)  (x  -  3) 

12.    (p  -  7.36)  (p  +  1.11) 

5. 

(2  X  +  7)  (x  -  3)            13.    (A  -  105)  (5  -  A) 

6. 

(2  X  -  7)  (3  X  - 

1)         14.    iX-{-nx){X  -2x) 

7. 

(1  -  x)  (2  -  X) 

15.   (A  -  5  a)  (A  +  a) 

8. 

(5  -  a)  (2  +  a) 

16.    (10  +  5  x)  (11  +  2  x) 

17.    (2.75  X- 

■f  1.22)  (5.08  x  + 1.55) 

18.    (10.33  X 

+  10.12  y)  (1.073  X  -  2.224  y) 

19.    (7.86  X  +  8.44)  (8.44  x  +  7.86) 

20.  (8  -  x)  (5  -  2/)  25.   (28  a  +  5  6)  (28  a  -  5  6) 

21.  (x  +  1.732)  (x  -  1.732)     26.   (70  p  -  1)  (70  p  -  1) 

22.  (a  -  72)  (a  -  101)  27.    (1  -  .037  x)  (1  -  .037  x) 

23.  {a  -  72)  (a  -  72)  28.    (7  -  12  xY 

24.  (a  -  101)  (a  -  101)  29.    (108  a  +  102  hy  , 

30.    (1.11  X  -  2.03  yy 

Division 

135.  In  division,  we  are  given  one  factor  of  a  number  to 
find  the  other.  The  given  factor  is  the  divisor,  the  required 
factor  is  the  quotient,  and  their  product  is  the  dividend.  The 
signs  of  the  separate  terms  of  the  quotient  must  be  such  that, 
when  those  terms  are  multiplied  by  the  terms  of  the  divisor, 
they  produce  the  signs  given  in  the  dividend. 

Express  the  four  possible  combinations  of  signs  for  sepa- 
rate terms  in  tabular  form,  and  show  that  from  the  table  one 
may  derive  the  rule: 

In  division^  like  signs  give  plus,  and  unlike  signs  minus. 


jm- 

EXERCISE  43 

Divide: 

1.   14x2by2aj 

9. 

-1.032x2  by  - 1.110  X 

2.   24a;5by4a;2 

10. 

-783x2  by  524  X 

3.   24  x^  by  4x2 

11. 

7.77  X  by -1.23  X 

4.   9x9  by  3x3 

12. 

2.346  by  -3.132 

5.    1056  x2  by  11  X 

13. 

4.836  X  by  3.142 

6.   28x2  by  -28  X 

14. 

-2.718x2  by  3.142x2 

7.    115  by  -5 

15. 

5.43  A  by  -3.45 

8.    -708  a  by  2  a 

16. 

7.08x3  by  -3.22  X 

17.   .67  x^  by  .37  X 

Long  Division 

136.    Model  F.  —  Divide  6x2  +  lo  -  19a;  by  3x  -  2. 

The  correspondence  of  multiplication  and  division  may  be 
seen  as  follows.  The  arrangement  here  shown  for  division 
is  recommended  as  the  most  compact.    ; 


Divisor 

3x 

-2 

Multiplicand 

3a: 

-2  ' 

Quotient 

2x 

-5 

Multiplier 

2x 

-5 

Dividend 

6a:2 

-19  a: 

+  10 

1st  partial  prod. 

6a:2 

-  4a: 

1st  subtrahend 

6x2 

-4  a: 

2d  partial  prod. 

-15x 

+  10 

-15  a: 

+  10 

Entire  product    6  x^ 

-19  a: 

+  10 

2d  subtrahend 

-15  a: 

+  10 

Check:    (-3) 

-(1)  = 

-3 

A  very  good  check  for  division  is  to  multiply  the  quotient 
by  the  divisor;  this  gives  a  set  of  partial  products  different 
from  the  subtrahends  obtained  in  dividing. 

EXERCISE  44 
Divide: 

1.  x2  -  17x  +  30by  X  -  2 

2.  22  +  9/i-/i2by  11  -/i 


1 18  TRANSFORMATIONS 

3.  9  2/2  _  49  by  3  2/  +  7 

4.  30  s2  +  65  s  +  10  by  6  s  +  1 

6.  60  s2  +  92st-\-S5t^hyl0s  +  7t 

6.  x^  -  1000  X  +  999  by  a;  -  1 

7.  240  A2  +  214  A  -  63  by  8  A  -  9 

8.  144  Q2  _  295  QX  +  50  X2  by  9  Q  -  10  X 

9.  504  m2  -  180  r2  -  1049  mr  by  56  m  +  9  r 

10.  9600  a"  -  2541  f  +  7436  at  by  300  a  -  77  t 

11.  3  m3  -  11  m2  -  4  m  +  30  by  m  -  3 
^12.   9  2/3  +  3  2/2-22/  +  30by32/  +  5 

13.  12  2^  _  11  ^3  _  4  ^2  _|_  25  0  -  14  by  3  2;  -  2 

.      14.  x'-6x^-\-5x''-x-12hyx''-2x  +  S 

>i&.  10  x4  +  29  x3  -  22  a;2  +  17  X  -  2  by  5  a;2  -  3  X  +  2 

16.  7  a:^  +  9  a^  -  37  a;2  +  41  a;  -  40  by  7  ^2  -  5  a;  +  8 


CHAPTER   VI 
IDENTITIES  AND  THEOREMS;  FACTORING 

137.  Compare  the  following  equations: 

I.   dx-\-  5  ==2x  +  7.  II.   3  (x  +  5)  =  3  X  +  15. 

Notice  that  while  the  first  equation  is  true  only  for  the 
particular  value  x  =  2,  the  second  is  true  for  any  value  that 
may  be  chosen. 

Again,  notice  that  the  expression  3  a;  +  5  cannot,  by  any 
means  we  know  of,  be  transformed  into  2  x  +  7;  while 
3  (x  +  5)  is  in  a  simple  way  transformed  into  Sx  -\-  15. 

138.  These  two  kinds  of  equations  have  distinct  names. 
An  equation  which  is  true  only  on  condition  that  the  let- 
ters in  it  have  particular  values  is  called  an  equation  of 
condition;  while  an  equation  which  is  true  for  any  values 
whatever  of  the  letters  in  it  is  called  an  identical  equation, 
an  equation  of  identity,  or  simply  an  identity. 

Equations  of  condition  are  the  equations  ordinarily  met 
with  in  solving  problems. 

Identical  equations  may  be  recognized  by  the  fact  that 
one  member  can  be  transformed  to  the  identical  form  of  the 
other  member.  The  two  members  are  said  to  be  identically 
equal,  and  sometimes  the  sign  =  is  used  instead  of  =  be- 
tween the  members  of  the  equation. 

The  Proof  of  Theorems 

139.  A  theorem  is  a  general  statement  requiring  proof.  In 
algebra,  a  theorem  is  often  proved  by  changing  the  statement 
to  an  equation,  and  showing  that  one  member  can  be  trans- 

119 


120         IDENTITIES  AND  THEOREMS;  FACTORING 

formed  so  as  to  become  exactly  like  the  other  member;  that 
is,  by  showing  the  equation  to  be  identical. 

140.  Theorem  A.  //  the  sum  of  any  two  numbers  is  multi- 
plied by  their  difference,  the  product  is  the  difference  of  their 
squares. 

Model  A.  —  Proof.  Let  a  and  b  represent  any  two 
numbers.  Then  the  theorem  is  expressed  by  the  following 
identity; 

(a  +  6)  (a  -  6)  =  a2  -  b^ 

The  first  member  is  transformed  into  the  second  by  a  simple  multi- 
plication. 

141.  Prove  the  following  theorems  i 

1.  The  square  of  the  sum  of  any  two  numbers  is  equal  to  the 
square  of  the  first  number,  plus  twice  the  product  of  the  two, 
plus  the  square  of  the  second. 

2.  The  square  of  the  difference  of  any  two  numbers  is  equal 
to  the  square  of  the  first  number,  minus  twice  the  product  of 
the  two,  plus  the  square  of  the  second. 

EXERCISE  46 

By  means  of  these  theorems  write  the  following  products 
at  sight,  without  performing  the  multiplications: 

1.  (a  +  3)2  3.   (a  +  132.1)2        5.   (a  -  7)^ 

2.  (a  +  7)2  4.   (a  -  3)2  e.   (a  -  132.1)2 

7.  (a  +  1)  (a  -  1)  10.    (a  +  38.7)  (a  -  38.7) 

8.  {a  +  3)  (a  -  3)  ii.    {a  +  xY 

9.  (a  +  7)  (a  -  7)  12.    (a  -  xf 

By  means  of  the  same  theorems  write  quotients  for  the 
following,  without  performing  the  divisions: 

13.  (a2  -  16)  -T-  (a  -  4) 

14.  (a2  -  121)  -^  (a  +  11) 


QUADRATIC  PRODUCTS  121 

15.  (a2~  10  a +  25)  -^  (a  -  5) 

16.  (a2  +  10  a  +  25)  -r-  (a  +  5) 

17.  (a2  +  100  a  +  2500)  ^  (a  +  50) 

18.  (a2  -  3600)  ^  (a  -  60) 

By  means  of  the  same  theorems  write  the  factors  of  the 
following  products: 

19.  x2  -  a2  25.  A2  -  81 J52 

20.  49  -  x2  •     26.  49  a2  -  4  A^ 

21.  4  -  4x  +  a;2  27.  B2  -  62 

22.  a;2  -  20  x-\-  100  28.   p2  _  28  p  +  196 

23.  x'  +  SOx^  1600  29.   256  +  32  n  +  n^ 

24.  a;2  -  120  a;  +  3600  so.    1  -  2  a  +  a^ 

By  Theorem  A  find  the  difference  of  the  squares  of  each 
of  the  following  pairs  of  numbers: 

31.  3  and  73  34.   339  and  319       37.  8133  and  8131 

32.  9  and  109         35.    1723  and  277      38.   2731  and  269 

33.  575  and  425     36.   121  and  120        39.   101  and  99 

40.    10001  and  1 

Quadratic  Products 

142.  The  study  of  the  product  of  two  factors  of  the  first 
degree,  like  (x  —  5)  (a?  +  2),  leads  to  certain  methods  of 
factoring  that  are  of  the  greatest  use  in  the  solution  of  an  im- 
portant kind  of  algebraic  equation. 

An  expression  in  which  there  is  a  term  of  the  second  de- 
gree, but  no  higher  term,  is  called  a  quadratic  expression. 
Consequently  products  like  {x  —  5)  (x  +  2)  are  called 
quadratic  products.  Equations  in  which  they  occur  are 
called  quadratic  equations. 


122       IDENTITIES  AND  THEOREMS;  FACTORING 


-  \ 


10 


143.   Model   B.  —  The  diagrams  on  this  page  are  recom- 
mended for  quadratic  products,  because  by  their  use  one  gets 
a  better  idea  of  the  way  in  which  such  products 
are  formed. 

The  products  of  terms  directly  under  each 
other  are  called  straight  products.  The  products 
of  terms  diagonally  opposite  each  other  are  called 
cross  products. 

The  straight  products  are  not  similar,  and  so 
cannot  he  united;  the  cross 
products  are  similar,  and  can 
he  united. 

In  this  example,  the  straight 
products  are  x^  and  — 10. 
The  cross  products  are  —  5x 
and  +  2  X,  and  their  sum  is 
—  S  X.  The  entire  product 
is  x2  -  3  X  -  10. 

144.  The  same  method  can 
be  applied  to  the  multiplica- 
tion of  numbers  of  two  fig- 
ures. For  example,  73  may 
be  expressed  as  7  i  +  3  and  78  as  7  i  +  8,  where  t  stands  for 
ten.  Then  the  straight  products  would  be  49  f  and  24, 
and  the  cross  products  21  t  and  56  t,  with  the  sum  77  t. 
Since  t  =  10,  the  straight  products  give  4924,  and  the  cross 
products  770,  the  sum  being  5694. 


EXERCISE  46 

Multiply  the  following  numbers  by  the  method  of  Section 
144: 

1.  34X37     3.  45X25      5.  36X45     7.  83X87      9.  28X73 

2.  58X52     4.  38X58     6.  35X65     8.  34X51     lo.  39X96 


QUADRATIC  PRQDUCTS 


123 


In  the  following  examples,  name  the  straight  products,  the 
cross  products,  the  sum  of  the  cross  products,  and  the  entire 


•odu 
11. 

ct: 

(x  -\-2){x-  7) 

26. 

(11  -h){2  +  h)     i 

12. 

(a  +  5)  (a  +  9) 

27. 

(4  6  -  1)  (2  6  -  3) 

13. 

(a  +  3)  (a  -  10) 

28. 

(2x  +  S){2x  +  3) 

14. 

(b  +  7)  (6  -  6) 

29. 

(31/ -7)  (3  2/ +  7) 

15. 

{h-\-S)(h  +  S) 

30. 

(6  s  +  1)  (5  s  +  10) 

16. 

(h  -  11)  {h  +  2) 

31. 

(x  +  2y){x-7y) 

17. 

{k  -l){k-  4) 

32. 

(a  +  5  6)  (a  +  9  &) 

18. 

(x  +  10)  (x  +  10) 

33. 

(3a +  26)  (5a +  26) 

19. 

(y  -  3)  (2/  +  3) 

34. 

(c  +  x)  (3  c  +  2  a;) 

20. 

(s  +  1)  (s  +  99) 

35. 

(3  /b  -  4  ^)  (5  /b  +  sf) 

21. 

(2  a;  +  1)  (7  a;  -  1) 

36. 

(lOx-2/i)  (3x+5/i) 

22. 

(5  a  +  1)  (9  a  +  1) 

37. 

(46-3/b)(36-2fc) 

23. 

(3a +  2)  (5a -2) 

38. 

(3:c  +  52/)(3x  +  5i/) 

24. 

(c  +  1)  (2  c  +  3) 

39. 

(3  2/-ll2)(3  2/+ll^) 

25. 

(/b  -  7)  (5  fc  +  1) 

40. 

(6  s  +  5  0  (10  s  -  7  0 

Factoring  by  Cross  Multiplication 

145.   Model   C.  —  In   the   product   x^  -  Q  x  -\-  U,   the 

straight    products   are   x^   and    14,    and    the  sum   of    the 

cross  products  is   —9x.     The  terms  that  give 

*^X^     the   straight   products   must   have    like   signs, 

^^  and   therefore   the   cross   products  must   have 

like    signs;     both    cross    products,    then,    are 

minus. 

To  get  the  first  straight  product,  we  have 

xXx;  to  get  the  second,  2X7;  then  the  signs  must  be 

chosen  so  that  both  cross  products  are  minus;  that  gives  us 

the  factors  (x  —  2)  (x  —  7). 


>«^^^ 
^^.>^ 


124       IDENTITIES  AND  THEOREJ^S;  FACTORING 


146. 


or  4  and  6. 
be  negative. 


Model  D.  —  In  the  product  x^  —  5  x  —  24,  the 
straight  products  are  x^  and  —  24;  —  5  x  is  the 
sum  of  the  cross  products.  The  terms  giving 
x^  for  a  product  had  hke  signs  and  the  terms 
giving  —24  for  a  product  had  unUke  signs; 
then  the  two  cross  products  had  unlike  signs, 
and  the  minus  cross  product  was  the  greater. 
To  get  x^  for  a  straight  product,  we  should 
have  to  multiply  x  and  x.  To  get  24,  we 
could  multiply  1  and  24,  2  and  12,  3  and  8, 
In  each  case,  the  larger  number  would  have  to 


-24 


Trjring  each  in  succession,  we  find  that  the  pair  of  factors 
{x  +  3)  (x  —  8)  gives  the  right  cross  products,  so  we  need 
not  try  farther. 

EXERCISE  47 

Name  the  straight  products  and  the  sum  of  the  cross  prod- 
ucts in  the  following  expressions,  and  find  their  factors: 

1.  x2  +  5x  +  6  6.  x^  +  5x  +  4 

2.  x2  -  2  X  +  1  7.  x2  +  7  X  +  12 

3.  x2  -  3  X  +  2  8.  x2  -  7  X  +  6 

4.  x2  -  4  X  -f  3  9.  x2  +  7  X  +  10 

5.  x2  +  4  a;  +  4  lo.  a?'  -  8  x  +  15 


QUADRATIC  PRODUCTS  125 

11.  x^  -Sx-{-7  41.  a;2  -  4 a;  -  60 

12.  x2  +  8  a;  +  15  42.  a;2  +  5  X  -  84 

13.  a;2  -  20  a;  +  19  43.  x^  -  5  x  -  150 

14.  x2  -  20  X  +  64  44.  a;2  +  6  X  -  40 

15.  a;2  +  20  a;  +  36  45.  a;^  -  6  x  -  91 

16.  a;2  +  20  a;  +  51  46.  x^  -  7  x  -  30 

17.  a;2  -  20  X  +  75  47.  a;2  -  6  X  -  55 

18.  a;2  +  20  x  +  84  48.  a;^  +  7  a;  -  78 

19.  a;2  -  14  a;  +  49  49.  a;^  +  7  x  -  170 

20.  x2  -  14  X  +  13  50.  a;2  -  8  x  ---  20 

21.  a;2  -  14  X  +  33  6i.  x^  -  9  x  -  10 

22.  x2  -  14  X  +  24  62.  x2  +  9  X  -  22 

63.  x2  -  10  X  -  200 

64.  x2  +  10  X  -  75 

65.  x2  +  10  X  -  96 

56.  X2  -  X  -  110 

67.  x2  +  2  X  -  120 

68.  x2  -  3  X  -  130 

59.  x2  +  X  -  132 

60.  x2  -  4  X  -  165 

61.  2x2  -  X  -  1 

62.  3  x2  -  2  X  -  1 

63.  3  x2  +  4  X  +  1 

64.  2x2  +  7x  +  6 

65.  2  x2  -  5  X  +  3 

66.  2  x2  -  X  -  3 

67.  3  x2  -  X  -  10 

38.  x2  -  4  X  -  32  68.  2  x2  +  3  X  -  9 

39.  x2  +  3  X  -  70  69.  5  x2  -  9  X  -  2 

40.  x2  _|_  4  a;  _  gg  70.  6  x2  +  X  -  15 


23. 

x2  +  14  X  +  45 

24. 

x2  +  14  X  +  48 

25. 

x2  -  8  X  +  16 

26. 

x2  +  17  X  +  60 

27. 

x2  -  19  X  +  90 

28. 

x2  -  18  X  +  32 

29. 

x2  +  18x.+  45 

30. 

x2  +  18  X  +  56 

31. 

'X2  -  X  -  2 

32. 

x2  -  X  -  6 

33. 

x2  -  3  X  -  4 

34. 

x2  +  X  -  6 

35. 

x2  +  2  X  -  15 

36. 

x2  +  3x-28 

37. 

x2  -  4  X  -  21 

126        IDENTITIES  AND  THEOREMS;   FACTORING 

71.  10x2  +  7a;+ 1  81.  9x2+17x-2 

72.  10  a;2  -  X  -  3  82.  6  x2  -  X  -  12 

73.  lOx^  -  ISx  -  3  83.  6 a;2  +  21  X  -  12, . 

74.  10  x2  -  17  X  +  3  84.  7  x2  -  9  X  -  10 

75.  3  x2  -  4  X  -  7  85.  11  x2  +  27  X  +  10 

76.  3  x2  +  22  X  -f  7  86.  15  x2  -  7  X  -  2 

77.  5  x2  +  8  X  -  4  87.  14  x2  +  11  X  -  15 

78.  5  x2  -  11  X  -  12  88.    21  X2  +  X  -  10 

79.  5  x2  +  28  X  +  15  89.    10  x2  -  29  X  -  21 

80.  10  x2  -  9  X  -  9  90.   6  x2  +  17  X  -  14 

Factor  the  following  expressions: 

91.  8x2  -  2  gg  1728x2  -  12 

92.  18  x2  -  50  99.  36  xY  -  25  26 

93.  75  a  -  27  a^  100.  100  h""  -  36  k^ 

94.  44  x3  -  275  xY  101.  1210  a^¥  -  10  ah 

95.  8  x2  -  162  102.  243  xyh"^  -  12  a^yz^ 

96.  18  x2  -  32  103.  1  -  100  «2 

97.  27  x2  -  147  104.  75  x^o  -  48  a^ 

105.   9  a^¥c^  -  9  x^e 

Completing  the  Square 
147.   Although  the  expression  (x  —  5)^  —  9  could  be  sim- 
plified and  factored  as  follows: 

(x  -  5)2  -  9  =  x2  -  lOx  +  25  -  9  =  x2  -  lOx  +  16 

=  (x  -  8)  (x  -  2) 

there  is  an  advantage  in  having  it  in  its  first  form,  because  by 
Theorem  A  we  can  at  once  factor  it  into 

(x  -  5)  -  3  and  (x  -  5)  +  3 

Then  simplifying  the  factors,  we  get  (x  —  8)  (x  —  2). 
The  advantage  is  greater  if  the  number  is  large,  as  in  the 
expression  (x  —  42)2  —  3^^   ^ 


COMPLETING  THE  SQUARE  127 

EXERCISE  48 
Factor: 

1.  {x  -  18)2  _  49  ^^  (^  _|.  90)2  _  36 

2.  (x  -  20)2  -  121  8.  (n  +  101)2  _  625 

3.  (a  -  32)2  _  100  9.  (25  -  xY  -  25 

4.  (2  a;  -  5)2  -  144  10.  (x  -  38)2  _  225 

5.  (x  -  35)2  _  81  11.  (3  a;  +  51)2  _  144 

6.  (p  +  28)2  -  900  12.  (m  -  59)2  -  400 

148.  In  factoring  an  expression  like  x^  -\-  5Q  x  -\-  768,  an 
easier  method  than  finding  all  pairs  of  factors  of  768  and 
selecting  the  right  pair  is  to  reduce  the  expression  to  the 
form  of  the  difference  of  two  squares.  We  may  then  apply 
Theorem  A  to  it,  as  in  Exercise  48. 

Taking  the  first  two  terms,  and  remembering  Theorem  1  of  Section 
141,  we  can  say  that  x^  +  56  x  is  a  part  of  a  perfect  square,  the  square 
of  some  binomial. 

Since  x^  is  the  square  of  the  first  term  of  the  binomial,  then  the  first 
term  of  the  binomial  must  be  x. 

Again,  56  x  is  twice  the  product  of  both  terms,  so  28  x  is  that  product; 
and  since  x  is  the  first  term,  28  must  be  the  second  term. 

Consequently,  a;^  +  56  a;  is  a  part  of  (x  +  28)^  =  x^  +  5Qx  +  784. 

149.  Model  E.  —  Factor  x^  +  5Q  x  +  768. 

x^  +  6Qx  +  7Q8  =  x^  +  5Qx+  (28)^  -  (28)2  +  768 
=  x2  +  56  re  +  784  -  784  +  768 
=  (x  +  28)2  -  16 
=  (x  -\-  28)2  -  (4)2 
=  (a:  +  28  +  4)  (a;  +  28  -  4) 
=  (a;  +  32)  (x  +  24) 

150.  The  process  by  which  we  get  a  perfect  square  from 
the  first  two  terms  of  our  given  expression  is  called 
completing  the  square.     It  may  perhaps  be  more  clearly 


128       IDENTITIES  AND  THEOREMS;  FACTORING 


understood  if  we  consider  the  diagram  for  cross  multiplica- 
tion: 


Since  the  middle 
term  is  a  double  cross 
product,  each  cross 
product  is  half  of  it. 


EXERCISE  49 
Factor  by  completing  the  square: 

1.  x'^-\-2x  -  4S 

2.  a2  -  28  a  +  187 

3.  x2  -  4  X  +  3 

4.  p2  _  36  p  +  128 

6.  A2  -  50  A  +  264 

6.  ^2  +  10  a:  -  24 

7.  x2  -  12  X  -  45 


11.     X' 


8.  n2  -  40  71  -  500 

9.  x^  -  6x  +  5 
10.   x2  +  10  a;  -  96 

21.    x^ 


2  -  62x  +  945 

12.  Z2  +  68X+  1155 

13.  s2  -  10  s  -  704 

14.  r2  -  50  r  +  616 

15.  t^  -12t-  1260 

16.  x^  -  112  X  +  3100 

17.  a2  +  204a  +  404 
18.^  Q^  +  78Q  +  1400 

19.  h^  -S2h  +  1561 

20.  x2  -  58  X  -  759 
72  X  +  1280 


151.  The  process  of  completing  the  square  becomes  some- 
what more  difficult  when  the  coefficient  of  the  middle  term  is 
an  odd  number. 

Model  F.  —  Factor  x^  +  11  x  -  726. 

a:2  +  11  a;  -  726  =  a;2  +  11  x  +  (5.5)2  _  30.25  -  726 
=  a;2  +  11  a;  +  30.25  -  756.25 
=  (x  4-  5.5)2  _  (27.5)2 
=  (x  +  5.5  +  27.5)  {x  +  5.5  -  27.5) 
=  (x  +  33)  {x  -  22) 


COMPLETING  THE  SQUARE  129 

The  method  of  completing  the  square  should  be  used  when 
the  factors  cannot  be  readily  seen  by  inspection. 

EXERCISE  50 
Factor  by  completing  the  square: 

1.  x'  +  x-Si  11,  x^  +  5x  -  594 

2.  x^  +  x-\-  t\  12.  x^  -9x-  792 

3.  x^  +  Sx-{-  1.25  13.  x2  -  15  X  -  1134 

4.  x^-\-  5x-{-2i  14.  x^  -  51x-\-  648 

5.  a;2  +  7  X  +  3  .25  15.  x^  +  81  x  +  1458 

6.  a2  +  11  a  -  5.75  le.  x^  -  75  x  +  1386 

7.  A2  -  13  A  +  6.25  17.  x2  +  87  X  +  1782 

8.  /b2  -  9  fc  +  4.25  18.  a;2  +  67  X  +  1120 

9.  X2  +  15  X  +  7.25               19.  x2  -  79  X  +  1350 
10.  Q^  +  3Q-  897f                 2o.  a^  +  83  a  +  1512 

21.  x2  -  11  X  -  2142 


CHAPTER  VII 
QUADRATIC   EQUATIONS 

152.  One  of  the  most  important  axioms  in  elementary 
algebra  is  the  following: 

The  product  of  two  or  more  factors  can  never  he  zero  unless 
at  least  one  of  these  factors  is  itself  zero. 

This  axiom  will  hereafter  be  referred  to  as  Axiom  A. 

153.  In  the  equation  2x  —  7  =  0,  x  must  have  the  value 
^;  if  we  substitute  this  value  for  x,  the  expression  2  a;  —  7 
becomes  zero. 

6  a;2  -  5  X  -  6  factors  into  (3  a;  +  2)  (2  x  -  3). 
To  make  3  x  +  2  =  0,  we  must  have  x  =  —  f . 
To  make  2  x  —  3  =  0,  we  must  have  x  =  |. 

EXERCISE  51 

What  value  must  we  substitute  for  x  in  each  factor  of  the 
following  expressions  in  order  that  the  factors  may  severally 
become  equal  to  zero? 

1.  x2-3x  +  2  6.   14x2  +  5x-l 

2.  x2-5x  +  4  7.   21x2-17x  +  2 

3.  6x2 -X- 2  8.   3x2-14x  +  16 

4.  2  x2  -  3  X  +  1  9.    12  x2  -  7  X  +  1 
6.  x2  -  X  -  2  lo.   15  x2  ~  11 X  -  14 

Two  Answers  to  One  Question 

154.  Model  A.  —  A  square  box  7  inches  high  has  160 
square  inches  more  in  its  lateral  surface  than  in  its  bottom. 
What  is  the  size  of  the  bottom? 

130 


TWO  ANSWERS  TO  ONE  QUESTION  131 

Let     X  =  the  number  of  inches  in  one  side  of  the  bottom. 
Then  x^  =  the  number  of  square  inches  in  the  bottom. 
7  X  =  the  number  of  square  inches  in  one  side. 

®     28  x  -  a:2  =  rBO 


(D 

0  =  x2  -  28  X  +  160 

®  +x2  -28  a; 

® 

0  =  (x-S){x  -  20) 

@  factored 

® 

0  =  x-8 

®  by  Ax.  A 

® 

S  =  x 

® +  8 

® 

0  =  X  -  20 

®  by  Ax.  A 

® 

20  =  X 

■    ®  +  20 

Ans.     8  in.  square,  or  20  in.  square. 

Check:  Area  of  bottom  =  64  sq.  in.  or  400  sq.  in. 
Area  of  1  side  =  56  sq.  in.  or  140  sq.  in. 
Area  of  4  sides  =  224  sq.  in.  or  560  sq.  in. 

224  -  64  =  160  560  -  400  =  160 

155.  This  example  has  two  answers,  or,  as  we  sometimes 
say,  there  are  two  values  of  x  which  will  satisfy  the  equation. 
These  two  numbers  are  technically  called  the  roots  of  the 
equation,  using  the  word  root  in  a  sense  entirely  different 
from  that  of  Sections  52  and  119. 

The  two  answers  do  not  mean  that  the  quantity  repre- 
sented by  X  can  actually  have  two  different  values  at  the 
same  time;  that  is,  the  same  box  cannot,  of  course,  be  8 
inches  square  and  also  20  inches  square.  But  a  box  8 
inches  square  would  have  the  properties  described  in  the 
equation,'  and  so  would  a  box  20  inches  square.  There 
are  two  sizes  of  square  box  that  could  be  made  so  as  to 
be  7  inches  high  and  at  the  same  time  so  as  to  have  160 
square  inches  more  in  the  lateral  surface  than  in  the 
bottom. 

When  an  equation  is  so  arranged  that  all  its  terms  are 
in  one  member,  and  the  other  member  is  zero  —  in  other 
words,  so  that  we  have  an  algebraic  expression  equated  to 
zero  —  then  the  factors  of  that  expression  are  sometimes 
loosely  called  the  "factors  of  the  equation." 


132  QUADRATIC  EQUATIONS 

156.  If  we  replace  by  a  the  number  7  in  the  problem  of 
Model  A,  and  by  h  the  number  160,  we  can  substitute  the 
following  numbers  for  a  and  h,  and  obtain  in  each  case  a 
new  problem. 


(a) 

5          3          4 

6 

10 

(&) 

36        20        39 
EXERCISE  52 

44 

175 

1.  In  a  square  storage  tank  SJ  ft.  deep,  th^e  are  32  sq.  yd. 
more  of  concrete  surface  in  the  sides  than  in  the  bottom. 
How  big  is  the  tank? 

2.  At  the  back  of  a  square  building  lot,  a  rectangular 
piece  of  ground  of  the  same  width  as  the  lot  and  containing 
2000  sq.  ft.,  is  added.  This  makes  the  lot  90  ft.  deep. 
What  is  its  frontage? 

3.  A  ventilator  opening  15  in.  high  gives  36  sq.  in.  more 
space  than  a  square  opening  of  the  same  width.  How  wide 
is  it? 

4.  I  bought  a  rectangular  lot  of  land,  containing  3850 
sq.  ft.  There  was  a  billboard  fence  along  the  front,  but  the 
other  three  sides  required  204  ft.  of  fencing.  What  was  the 
depth  of  the  lot? 

6.  A  man  contracted  to  pay  15  cents  a  foot  for  gilt  picture 
molding  and  20  cents  a  square  yard  for  straw  matting  used 
in  furnishing  a  square  room  in  his  house.  He  was  astonished 
to  find  that  the  picture  molding  cost  him  $4  more  than  the 
matting.     What  size  was  the  room? 

6.  Another  room  was  2  yd.  longer  than  it  was  wide. 
For  this  room,  the  molding  at  15  cents  per  foot  cost  $4.20 
more  than  the  matting  at  20  cents  per  square  yard.  What 
were  the  dimensions  of  the  floor? 


TWO  ANSWERS  TO  ONE  QUESTION  133 

7.  There  is  a  cylinder  8  in.  in  height  (h),  whose  lateral 
surface  (2  irrh)  exceeds  the  area  of  its  base  (ttt^)  by  88  sq.  in. 
Find  its  radius  (r).     (Use  tt  =  3j.) 

*  8.  A  man  sold  a  horse  for  $102  and  found  that  his  loss  per 
cent  was  one  eighth  of  the  number  of  dollars  he  had  paid  for 
the  horse.     How  much  had  he  paid? 

9.  In  the  formula  ¥  —  {a  —  xY  =  &  —  x^,  what  value 
must  a  have  if  x  =  60  when  6  =  87  and  c  =  65? 

10.  In  the  formula  of  Ex.  9,  if  a  =  28,  6  =  17,  and 
c  =  X  4-  5,  how  much  is  xl 

Use  the  following  numbers  to  make  new  problems  from 
Exs.  1-9. 


Example  1. 

(a  =  81;  h  =  32) 

(a) 
(Jb) 

6 
15 

9 
35 

7h 
24 

8 
61 

10 
421 

Example  2. 

(a  =  2000;  h  =  90) 

(a) 
(&) 

1200 
70 

420 
44 

360 
46 

2100 
100 

2400 
110 

Example  3. 
(a  =  15;  6  =  36) 

(a) 
ib) 

22 

72 

19 
60 

33 
175 

20 
91 

36 
180 

Example  4. 

(a  =  3850;  b  =  204) 

(a) 
(b) 

2240 
136 

3000 
160 

3920 
196 

4140 
226 

4400 
260 

Example  6. 
(a  =  15^;  b  =  20ff; 

c  =  $4) 

(a)        10^ 
(6)        15^ 
(c)  $1.05 

m 

$2.30 

8^         ISi 
18?^        36^ 
56^  $1.20 

25,i 

$1 
$2.00 

Example  6. 
(a  =  2yd.;  &  =  15^; 
c  =  20jf;  d  =  $4.20) 

(a)  2  ft 

(b)  10^ 

(c)  21^ 
(d)$1.05 

,.        2  ft 
7?i 
15^ 
$1.08 

2  yd.     2  yd.     3  yd. 
10?^        15^        11^ 

18^        27^         IQ^ 
$2.00    $3.00    $2.78 

Example  7. 

(a  =  8;  6  =  88) 

(a) 

(&) 

25 

154 

5 
66 

6 
110 

9 
176 

13 

418 

Example  8. 
(a  =  $102;  b  =  i) 

(a) 
(6) 

$144 

$147 

$105 

$192 

$105.60 

Example  9. 
(a;  =  60;  6  =  87;  c  =  65) 

(:^) 
(&) 
(c) 

37 
41 

47 

95 

70 

118 

66 
47 
79 

48 
46 
62 

10 
40 
38 

134  QUADRATIC  EQUATIONS 

Both  Answers  Alike 

157.  Model  B.  —  In  the  problem  of  Model  A  (p.  130),  if 
the  number  160  had  been  196,  both  answers  would  have  been 
alike.  The  problem  would  read:  A  square  box  7  inches 
high  has  196  square  inches  more  in  its  lateral  surface  than 
in  its  bottom.     What  is  the  size  of  the  bottom? 

Let      X  =  the  number  of  inches  in  one  side  of  the  bottom. 
Then  x^  =  the  number  of  square  inches  in  the  bottom. 
7  X  =  the  number  of  square  inches  in  one  side. 
©    28  X  -  a:2  =  196 

®  0  =  a:2  -  28  a;  +  196  ©  +  a;^  -  28  a; 

®  0  =  {x  -  U){x  -  14)       ©  factored 

©        X  -  U  =  0  ©  by  Ax.  A 

®  X  =  U  ©  +  14 

Ans.     Bottom  of  box,  14  in.  square. 
Check:       Area  of  bottom  =  196  sq.  in. 
Area  of  1  side  =    98  sq.  in. 
Area  of  4  sides  =  392  sq.  in. 
196  +  196  =  392 

Here,  on  account  of  a  special  selection  of  figures  (such 
as  might  accidentally  appear  in  any  problem),  the  two 
answers  are  alike.  Similar  results  are  obtained  if,  instead 
of  7  inches  and  196  square  inches,  we  say  9  inches  and  324 
square  inches,  or  5  inches  and  100  square  inches. 

EXERCISE  53 

Solve  the  problems  of  Exercise  52,  making  the  substi- 
tutions indicated  for  each  problem: 

1.  For  the  numbers  8 J  and  32  substitute  25i  and  289, 
respectively. 

2.  For  2000  substitute  2025. 

3.  For  15  and  36  substitute  16  and  64,  respectively. 

4.  For  3850  substitute  5202. 

5.  For  15 jz^  and  $4  substitute  lOj?^  and  $1.80,  respectively. 


/ 


ANSWERS  APPARENTLY  DIFFERENT  135 


6.  For  15j?f  and  $4.20  substitute  10^  and  $2.20,  respectively. 

7.  For  8  and  88  substitute  7  and  154,  respectively. 

8.  For  102  substitute  200. 

9.  For  87  substitute  25. 

10.  For  28,  17,  and  5  substitute  18,  14,  and  4,  respectively. 

Answers  Apparently  Different 

158.  If  we  find  two  different  numbers  as  the  values  of  x 
in  solving  a  quadratic  equation,  we  naturally  conclude  that 
this  indicates  two  answers  for  the  problem  from  which  the 
equation  was  formed.  We  sometimes  find,  however,  that 
the  two  answers  thus  indicated  are  actually  the  same. 

Model  C.  —  Divide  28  into  two  parts  whose  product 
shall  be  75. 

Let  X  =  one  part;  then  28  —  x  =  the  other  part. 

®  same  values 
@  +  a:2  -  28  a; 
(D  factored 
0  by  Ax.  A 
® +  25 
®  by  Ax.  A 
®  +  3 

=  3,  28  -  X  =  25. 

25  and  3,  or  3  and  25. 

The  first  answer  means  that  one  part  can  be  25  and  then  the  other 
must  be  3;  and  the  second,  that  one  part  can  be  3  and  then  the  other 
must  be  25.     The  two  answers  are  therefore  really  the  same. 
Check:    25  X  3  =  75;  25  +  3  =  28. 

EXERCISE  54 

1.  A  rectangular  field  containing  one  acre  requires  924 
feet  of  fencing.  What  are  its  dimensions?  (An  acre  is 
10  square  chains  and  a  chain  is  66  feet  long.) 


© 

X  (28  -  x)  =  75 

(D 

28  X  -  x2  =  75 

® 

0  =  x2  -  28  X  +  75 

® 

0  =  (x-  25)  (x  -  3) 

0 

0  =  X  -  25 

(D 

25  =  x 

® 

0  =  X  -  3 

® 

3  =  x 

® 

When  X  =  25,  28  -  X  =  3;  when  x 

Ans. 

136  QUADRATIC  EQUATIONS 

.  2.  Two  cubical  bins,  side  by  side,  extend  the  whole  length 
of  a  16-foot  wall,  and  will  exactly  contain  OJ  cords  of  kindling 
wood.  What  is  the  size  of  each?  (A  cord  of  wood  measures 
4  by  4  by  8  feet.) 

3.  The  number  30,551  has  two  factors  whose  sum  is  360. 
What  are  the  factors? 

4.  A  rectangular  room  requires  68  feet  of  picture  molding. 
The  same  room  requires  32  square  yards  of  carpet.  What 
is  the  size  of  the  room? 

5.  A  wharf  projects 3^7  feet  into  the  water,  and  is  made 
of  two  platforms;  its  total  area  is  377  square  feet.  What 
is  its  greatest  width  and  its  least? 

6.  Forty-six  rods  of  fencing  are  required  for  a  rectangular 
field  whose  diagonal  is  17  rods.     What  is  the  size  of  the  field? 

7.  A  rectangular  corner  lot  and  two  square  lots  adjoining 
it  contain  in  all  9972  square  feet,  and  have  altogether  114 
feet  frontage  on  each  street.  All  the  lots  on  a  street  have 
the  same  depth.     What  are  the  dimensions  of  the  lots? 

8.  By  one  pipe  a  certain  tank  is  filled,  and  then  by  an- 
other immediately  emptied,  the  whole  operation  requiring 
an  hour  all  but  ten  minutes.  Then  both  pipes  are  used 
to  fill  the  tank  again,  which  is  done  in  12  minutes,  How 
long  would  it  take  each  pipe  separately  to  fill  the  tank? 

The  Meaning  of  Negative  Answers 

159.  Sometimes  one  answer  is  negative  and  still  can  be 
interpreted  as  a  reasonable  answer.  It  is  generally  neces- 
sary to  change  the  sense  of  some  word,  so  as  to  get  a  meaning 
exactly  opposite  to  the  meaning  in  the  original  statement 
of  the  problem. 

Model  D.  —  Of  two  pipes,  one  takes  3  hours  longer  than 
the  other  to  fill  a  cistern,  and  both  together  would  fill  it  half 


THE  MEANING  OF  NEGATIVE  ANSWERS  137 

full  in  an  hour.    How  bng  would  it  take  each  pipe  alone  to 
fill  it? 


1 


Let  X  =  the  number  of  hours  for  one  pipe  to  fill  it. 

Then  x  +  S  =  the  number  of  hours  for  the  other  pipe. 

the  fraction  of  the  cistern  filled  by  the  first  pipe  in 

one  hour. 

the  fraction  filled  by  the  other  pipe  in  one  hour. 


X  +  3 


11          1 

® 

x^  {x  +  S)~  2 

® 

2x  +  Q-\-2x  =  x{x  +  S) 

(D 

Ax  +  6  =  x^  -\-Sx 

0 

0  =  x2  -  X  - 

© 

0  =  (a;  -  3)  0 

(i)X2x{x  +  S) 
@  same  values 
@  -  4x  -  6 
2)  ®  factored 

Whence  we  obtain  x  =  S  and  x  +  3  =  6;  that  is,  one  pipe  would 
fill  it  in  3  hours  and  the  other  in  6  hours. 

But  we  also  obtain  x  =  —2  and  x  +  3  =  1;  that  is,  the  first  pipe  is 
emptying  the  cistern,  and  by  itself  could  do  that  in  2  hours,  while  the 
second  would  fill  it  in  one  hour. 

Check:  The  first  pipe  fills  ^  of  the  cistern  in  one  hour;  the  second 
fills  i  of  it  in  one  hour;  both,  ^. 

Check:  The  first  pipe  takes  out  ^  the  cistemful  in  one  hour;  the 
second  pours  in  the  whole  capacity;  there  is  left  enough  to  fill  ^  the 
cistern. 

EXERCISE  55 

1.  A  field  20  rods  by  18  rods  is  made  352  square  rods  in 
area  by  cutting  off  a  strip  of  uniform  width  on  the  side,  and 
adding  a  strip  of  the  same  width  on  the  end.  What  is  the 
width  of  the  strips? 

2.  If  a  fraction  having  70  as  its  numerator  is  inverted 
and  multiplied  by  6,  its  new  value  will  exceed  its  original 
value  by  1.     Find  the  fraction. 

3.  A  boat  which  is  able  to  go  5  miles  an  hour  in  still 
water  goes  21  miles  southward  on  a  certain  river  and  then 
24  miles  northward,  the  whole  time  spent  in  travehng  being 
11  hours.     How  fast  does  the  stream  flow  southward? 


138  QUADRATIC  EQUATIONS 

4.  Both  pipes  of  a  certain  water  tank  are  opened,  and  the 
tank,  being  empty  at  noon,  is  filled  at  6  o'clock.  One  of 
the  pipes,  if  used  alone,  would  take  35  hours  longer  than  the 
other  pipe  alone  to  fill  the  tank.  At  what  time  would  the 
tank  be  full  in  each  case? 

5.  In  the  equation  c^  =  a^  +  b^,  if  a  =  2h  +  2  and 
c  =  13,  what  does  h  equal? 

6.  In  the  equation  of  Ex.  5,  if  6  =  a  —  2  and  c  =  58, 
what  does  a  equal? 

In  Exs.  1-4,  the  following  numbers  may  be  used  to  form 
new  problems: 


Example  1. 

(a) 

20 

16 

25 

32 

35 

(o  =  20;  6  =  18; 

(&) 

13 

10 

22 

30 

12 

c  =  352) 

(c) 

230 

105 

540 

880 

342 

Example  2. 

(a) 

8 

72 

16 

45 

80 

(o  =  70;  6  =  6; 

(h) 

4 

36 

144 

900 

16 

c  =  1) 

(c) 

-3 

5 

7 

11 

15 

Example  3. 

(a) 

3 

7 

6 

13 

8 

(a  =  5;  6  =  21; 

(&) 

12 

34 

30 

62 

22 

c  =  24;  d  =  11) 

(c) 

10 

33 

32 

63 

30 

id) 

8 

10 

19 

10 

8 

Example  4. 

(a) 

3 

6 

4 

2 

8 

(a  =  6;  6  =  35) 

(&) 

8 

5 

15 

3 

63 

Answers  Suggesting  Related  Problems 

160.  A  negative  answer  to  a  quadratic  equation  may  not 
be  interpretable  as  a  reasonable  answer  to  the  problem 
which  gave  rise  to  the  equation,  but  may  suggest  a  new 
problem,  closely  related  to  the  given  one,  and  having  the 
negative  and  positive  answers  interchanged.  Such  a  case 
is  the  following: 

Model  E.  —  A  farmer  bought  a  herd  of  cows  for  $400. 
If  there  had  been  4  more  in  the  herd,  the  price  would  have 
been  $5  less  apiece.    How  many  were  there? 


ANSWERS  SUGGESTING  RELATED  PROBLEMS      139 


Let 

X  =  the  number  of  cows  in  the  herd 

• 

Then 

400 

—  =  the  number  of  dollars  paid  for 

X 

400        400     ,    ^ 

one  cow. 

© 

X    -x  +  4  +  .^ 

® 

400  x  +  1600  =  400  X  +  5  a:2  +  20  a: 

©  X  X  (x  +  4) 

® 

0  =  5  x2  +  20  x  -  1600 

®  -  400  X  -  1600 

® 

0  =  rc2  +  4  a;  -  320 

©  -5 

© 

0  =  (x  -  16)  (x  +  20) 

®  factored 

© 

x  =  16;  X  =  -20 

©  by  Ax.  A 

Ans.     16  cows. 

Check:      16  cows  for  $400,     $25  apiece  ' 
4  more,  or  20  cows,  for  $400,  $20  apiece 

25  -  20  =  5 

Now  we  can  change  the  algebraic  sense  of  the  problem  by 
changing  words  like  "  more  "  to  ''less  "  and  vice  versa.  The 
new  problem  would  read:  A  farmer  bought  a  herd  of  cows 
for  $400;  if  there  had  been  4  less  in  the  herd,  they  would 
have  cost  $5  more  apiece.     How  many  were  there? 

Let  X  =  the  number  of  cows  in  the  herd, 

^     400        400 

®     = A  —  5 

^  X  X    —   4: 

Whence  x  =  20,  or  —16. 
Check:     20  cows  for  $400,     $20  apiece 
4  less,  or  16  cows,  for  $400,     $25  apiece 
25  =  20  +  5 


EXERCISE  56 

In  the  following  problems,  the  negative  answers  suggest 
related  problems  in  the  opposite  algebraic  sense. 

1.  There  are  40  rooms  in  a  house,  and  3  more  rooms  on 
each  floor  than  there  are  floors  in  the  house.  How  many 
floors  are  there? 

2.  Comparing  a  pail  with  a  140-quart  tub,  a  man  con- 
cluded that  the  tub  held  as  many  pailfuls  as  the  pail  held 


140  QUADRATIC  EQUATIONS 

quarts.     He  found,  however,  that  the  tub  held  4  pailfuls 
more.     What  was  the  capacity  of  the  pail? 

3.  In  order  to  save  4  hours  on  his  regular  trip  of  126 
miles,  a  stage  driver  finds  it  necessary  to  go  2  miles  an  hour 
faster  than  his  usual  rate.     What  is  his  usual  rate? 

4.  What  is  the  price  of  eggs  when  10  more  in  a  dollar's 
worth  lowers  the  price  4  cents  a  dozen? 

5.  A  rectangular  field  contains  IJ  acres.  If  its  length  is 
increased  by  20  feet  and  its  breadth  diminished  by  18  feet, 
its  area  will  be  diminished  2340  square  feet.  What  are  its 
dimensions? 

For  Exs.  1-5  the  following  numbers  may  be  used  to  form 


new  problems: 

Example  1. 

(a) 

35 

50 

60 

44    . 

112 

(a  =  40;  6  = 

3) 

(&) 

2 

5 

4 

7 

6 

Example  2. 

(a) 

88 

176 

99 

90 

180 

(a  =  140;  b  -- 

=  4) 

(6) 

3 

5 

2 

^ 

3 

Example  3. 

(a) 

6 

12 

4 

2 

5 

(a  =  4;  6  =  : 

126; 

(b) 

176 

135 

120 

180 

150 

c  =  2) 

(c) 

3 

4 

11 

1 

2i 

Example  4. 

(a) 

5 

2 

25 

50 

20 

(a  =  10;  6  = 

4) 

(&)' 

1 

1 

8 

2 

5 

Example  5. 

(a) 

2 

3 

h 

4 

2h 

(a  =  U;  6  = 

20; 

ib) 

24 

14 

12 

-60 

-45 

c  =  18;  d  = 

=  2340) 

(c) 

20 

30 

10 

-16 

-30 

id) 

3120 

7680 

780 

6240 

-3600 

Meaningless  Answers 

161.  Jn  each  of  the  following  problems,  an  answer  will  be 
obtained  which  satisfies  the  quadratic  equation,  but  has  no 
reasonable  interpretation  in  the  problem  which  gave  rise 
to  that  equation;  and  the  answer  may  not  even  suggest  a 
related  problem. 


MEANINGLESS  ANSWERS  141 

EXERCISE  57 

1.  A  square  box  7  inches  high  and  without  a  cover  has 
288  square  inches  of  inside  surface.  What  is  the  size  of  the 
bottom? 

2.  Of  two  large  wheels,  one  makes  48  turns  more  than  the 
other  in  roUing  a  mile.  Their  tires,  straightened  out  and 
laid  end  to  end,  would  reach  21  feet.  What  are  their  cir- 
cumferences? 

3.  Two  sprinters  start  in  opposite  directions  from  the 
middle  post  of  a  straightaway  440-yard  track.  In  one 
second  they  are  21  yards  apart,  and  one  of  them  finishes 
his  sprint  2  seconds  ahead  of  the  other.  What  is  the 
speed  of  each? 

4.  A  tank  containing  4  gallons  is  filled  by  one  pipe  and  then 
immediately  emptied  by  another,  both  operations  requiring 
just  5  hours;  one  pipe  dehvers  3  gallons  per  hour  more 
than  the  other.  How  many  gallons  can  each  pipe  deliver 
in  an  hour? 

5.  A  man  rows  5  miles  downstream  and  back  in  4  hours. 
The  stream  flows  3  miles  an  hour.  How  fast  can  the  man 
row  in  still  water? 

In  Exs.  1-5  the  following  numbers  may  be  used  to  form 


new  problems: 

Example  1. 

(a  =  7;  6  =  288) 

(a) 
(b) 

8 
185 

10 
176 

15 
^44 

20 
249 

3 

36i 

Example  2. 

(a  =  48;  b  =  21) 

(«) 
(6) 

40 
23 

10 
32i 

88 
22 

96 
241 

88 
27 

Example  3. 
(o  =  440;  b  =  21; 

c  =  2) 

(«) 
(6) 
(c) 

400 

18 

5 

300 
22 
21 

1000 
20i 
22^ 

900 
201 
10 

500 
241 

Example  4. 
(a  =  4;  6  =  5; 
,  c  =  3) 

(a) 
(c) 

6 
5 

.     7 

8 
7 
9 

10  ' 
3 

7 

4 
3 
5 

21 

13^ 

5 

142  QUADRATIC  EQUATIONS 


Example  6. 

(a) 

15 

24 

30 

25i 

10^ 

(a  =  5;  6  =  4; 

(&) 

9 

14 

12 

10 

5 

c  =  3) 

(c) 

4 

5 

6 

7 

2 

Equations  Involving  Irrational  Numbers 

162.  Model  F.  —  What  is  the  length  of  one  side  of  a  square 
whose  area  is  3  square  feet? 

Let  X  =  the  number  of  feet  in  one  side  of  the  square. 

©  a;2  =  3 

@       x2-3  =  0  ©-3 

This  last  equation  could  be  wTitten 

®     x2  -  (V3)2  =  0  ©  same  values 

In  equation  ©,  V3  is  the  symbol  for  a  number  that,  multiplied  by 
itself,  would  give  3  for  a  product. 

®     (x  +  V3)  {x  -  V3)  =  0     ®  factored 

163.  The  exact  value  of  \/3  cannot  be  written  in  ordi- 
nary figures.  The  square  of  1.7  gives  2.9,  the  square  of 
1.732  gives  2.999,  the  square  of  1.73205  gives  2.9999,  and 
so  on.  The  squares  are  nearer  and  nearer  to  3,  but  never 
exactly  equal  to  it.  Again,  the  numbers  If,  If,  ly\,  1|^, 
If  T>  give  squares  that  are  nearer  and  nearer  to  3,  but  never 
exactly  equal  to  it. 

Numbers  that  cannot  be  exactly  expressed  in  ordinary 
figures  are  called  irrational  numbers. 

164.  Whether  we  use  the  symbol  \/3,  which  represents 
the  exact  value  of  the  number  that  multiplied  by  itself  will 
give  3  for  a  product,  or  whether  we  use  a  more  or  less  ac- 
curate approximation  (such  as  1.7,  or  1.73,  or  1.732),  will 
depend  in  every  case  on  the  use  that  we  intend  to  make  of 
our  results. 

In  general,  if  the  results  are  to  be  used  for  arithmetical 
computation  or  for  measurement,  it  is  desirable  to  use  the 
approximate  value,  the  number  of  figures  being  chosen  ac- 
cording to  the  needs  of  the  problem.    If,  on  the  other  hand, 


EQUATIONS  INVOLVING  IRRATIONAL  NUMBERS      143 

we  intend  to  use  our  result  in  further  algebraic  work,  the 
algebraic  symbol  V3  may  be  found  more  convenient.  Sys- 
tematic methods  of  dealing  with  such  symbols  will  be  given 
in  later  years. 

165.  The  roots  of  the  equation  x^  =  3,  or  of  any  equation 
which  reduces  to  x^  =  some  number,  are  said  to  be  equal 
and  opposite.  That  is,  the  value  of  x  turns  out  to  be  either 
a  particular  number  or  its  negative.  Thus  the  equation 
x^  =  16  gives  X  =  4:  and  x  =  —4. 

Notice  that  the  word  root  means  an  entirely  different  thing 
in  the  following  sentence: 

There  are  two  square  roots  of  16  (or  of  any  number);  one 
is  +  and  the  other  — . 

166.  A  quadratic  equation  without  any  first-degree 
term  is  sometimes  called  a  pure  quadratic  equation. 

167.  We  cannot  factor  by  inspection  the  equation 

x2  -  10  X  +  20  =  0 

By  completing  the  square  of  which  a;^  —  10  x  are  the  first  two  terms, 
we  change  this  equation  into  the  form  of  the  difference  of  two  squares, 
and  obtain  (a;^  -  10  a;  +  25)  -  5  =  0 

which  can  be  written  {x  —  5)^  —  (V^)^  =  0 

and  factored  thus  (x  —  5  +  ^y5)  {x  —  5  —  \^5)  =  0 

Using  the  value  V5  =  2.236,  these  factors  become 
{x  -  2.764)  {x  -  7.236)  =  0      . 

EXERCISE  58 

Factor  the  following  expressions  and  find  the  approximate 
numerical  values  of  your  factors,  using  \/2  =  1.414, 
^3  =  1.732,  V5  =  2.236,  V6  =  2.450,  VIO  =  3.162. 

1.  (x  -  2)2  -  2         5.    (a  -  10)2  _  6       9.    (A  +  10)2  _  iq 

2.  (x  -  5)2  -  3         6.   (a  +1)2-2       10.   (16p  -  1)2  -  6 

3.  (x-7)2-5         7.    (5  +  a)2-5       ii.   (2a; -5)2 -3 

4.  (2x  -  dy  -  10    8.    ip  -  3)2  -  3       12.   (4m  +  7)2  -  5 


144  QUADRATIC  EQUATIONS 

Solve  the  following  equations: 

13.  a;2  +  6  a;  -  3  =  0  20.  2  A;^  -  24  A;  +  62  =  0 

14.  x2- 10a;+ 15  =  0  21.  4x2+16^-24=0 

15.  x2  +  2  X  -  5  =  0  22.  2  p2  +  12  p  +  8  =  0 

16.  A^  +  QA-1  =0  23.  x2  -  17.2  =  0 

17.  p2  _  24  p  +  141  =  0  24.  p2  +  3  p  _  8  75  =  0 

18.  A^  +  20^  +  98  =  0  25.  X2  +  5X  =  11 

19.  ^2  4-  30  ^  +  220  =  0  26.  a"  +  17.3  =  17.3  a 

27.   344  -  a:2  =  52.6  x 


168. 


Peculiar 

Quadratics 

Model  G. 

—  Solve  the  equation 

X' 

=  3x 

® 

x^  =  3x 

® 

a:2  - 

-  3a;  =  0 

(D-^x 

® 

X  {x 

-3)  =0 

@  factored 

0 

X  = 

0,    a;  =  3 

0  by  Ax.  A 

Notice  that  the  value  x  =  0  satisfies  the  equation.  This 
equation  is  said  to  have  one  null  root. 

169.   Model  H.  —  Solve  the  equation 

x2  =60:-  11 

Q  a;2  =  6a;-ll 

@  x2-6x  +  ll=0  ®-6a:  +  ll 

®     (a;2-6a;  +  9)  +  2  =  0  ®  same  values 

Since  ®  is  the  sum  of  two  squares,  we  cannot  factor 
it;  the  appearance  of  such  an  equation  shows  that  some 
absurdity  existed  in  the  statement  of  the  problem  that 
gave  rise  to  it.  The  equation  is  technically  said  to  have 
imaginary  roots.  Equations  of  this  type  will  be  studied 
in  later  years  of  the  mathematics  course. 


PECULIAR  QUADRATICS  145 

EXERCISE  59 

1.  Three  times  the  square  of  a  number  is  equal  to  10 
times  the  number  itself.     What  is  the  number? 

2.  A  room  is  twice  as  long  as  it  is  wide,  and  its  floor  con- 
tains 144 J  square  feet.     How  wide  is  it? 

3.  A  square  room  9  feet  high  takes  twice  as  much  paper 
to  cover  the  walls  as  to  cover  the  ceiling.  What  is  the  size 
of  the  ceiling? 

4.  In  a  current  of  2  miles  an  hour,  a  man  takes  8  hours 
longer  t9  row  24  miles  upstream  than  to  row  the  same  dis- 
tance downstream.     How  fast  can  he  row  in  still  water? 

5.  A  field  20  rods  by  18  is  made  353  square  rods  in  area 
by  cutting  off  a  strip  of  uniform  width  on  the  longer  side 
and  adding  a  strip  of  the  same  width  on  the  end.  What  is 
the  width  of  the  strips? 

6.  A  square  box  9  inches  high  has  300  square  inches  more 
on  the  sides  than  on  the  bottom.  What  are  the  dimensions 
of  the  box? 

7.  One  man  takes  4  hours  longer  than  another  to  saw  a 
cord  of  wood,  and  both  working  together  can  saw  it  in  3 
hours.     How  long  would  it  take  each  man  working  alone? 

8.  Two  wheels,  the  sum  of  whose  circumferences  is  equal 
to  8  yards,  roll  the  same  distance.  The  number  of  turns 
made  by  one  wheel,  added  to  the  number  of  turns  made  by 
the  other  wheel,  gives  a  sum  equal  to  the  number  of  yards 
in  this  distance.     Find  the  circumferences  of  the  wheels. 

Use  a  letter  for  the  number  of  yards  traversed;  also  one  other  letter. 

9.  A  line  100  centimeters  long  is  divided  into  two  parts, 
such  that  the  longer  part  is  contained  in  the  whole  line  as 
many  times  as  the  shorter  part  is  contained  in  the  longer 
part.    What  is  the  length  of  each  part? 


146  QUADRATIC  EQUATIONS 

10.  Across  the  back  of  a  square  building  lot,  a  rectangular 
piece  of  ground  containing  1820  square  feet  is  added.  The 
lot  is  then  75.3  feet  deep.     What  is  its  frontage? 

170.   Model  I.  —  Solve  the  equation 
23.4  a;2  + 508  X  =3420 


® 

23.4  x'  +  508  X  =  3420 

0 

23.4  x'  +  608  X  -  3420  =  0 

0  -3420 

® 

^  +  21.7  a;  -  146  =  0 

0  +  23.4 

0 

y?  +  21.7  X  +  (10.8)«  -  263  -  0 

0  same  values 

0 

{x  +  10.8)«  -  (16.2)2  „  0 

0  same  values 

0 

(x  4-  10.8  +  16.2)  (x  +  10.8  -  16.2)  =  0 

0  factore<l 

0 

(x  +  27.0)  (x  -  6.4)  =  0 

®  same  values 

0 

X  =  6.4,    X  -  -27.0 

0  by  Ax.  A 

Check: 

X  =  6.4                                X  -  -27.0 
x«  =  29                                x»  =  729 
23.4  a;^-    680                    23.4  x«  =     17100 
608X-2740                      608X--13700 

3420  3400 

A  less  laborious  check  is  to  multiply  the  factors  (x  —  6.4)  (x  -|-  27.0), 
which  gives  x*  +  21.6x  —  146,  sufficiently  doHe  to  0. 

Then  the  transformation  0  to  0  would  have  to  be  checked  by  mul- 
tiplying 23.4  X  21.7  and  23.4  X  146. 

EXERCISE  60 

1.  x^^\Z2x-  183  =0 

2.  x2-43.5aj4-3.77  =0 

8.   12.3  x2- 45.6  aj- 78.9  =0 
4.  38.4  -x2  =  31.8  a; 
38.7 


o. 

X  (X  4- 10.6)  "" 

6. 

1.37  x^^- 2.41  x- 3.59  =0 

7. 

2.83  X  «x2-34.7 

8. 

58.2 +  x2  =53.6x 

e. 

3.86  x+ 1.03x2  =873 

10. 

x(3.55x-11.9)  =67.8 

CHAPTER  vrn 

SOME  THEOREMS  OF  GEOMETRY 

171.  Geometry  originally  meant  land-measurement  The 
ancient  Egyptians  cultivated  lands  on  which  the  landmarks 
were  periodically  washed  away,  and  they  found  it  neces- 
sary to  measure  and  relocate  their  holdings.  Upon  their 
work  the  Greeks  based  a  wonderful  science,  which  is  grow- 
ing even  at  the  present  day.  Some  of  the  theorems  of  that 
science  were  stated  in  Chapter  III,  as  rules  for  measure- 
ment. Upon  them  depend  those  that  follow,  some  of  the 
most  important  and  fundamental  facts  in  geometry. 

172.  It  is  customary  to  represent  the  number  of  degrees 
in  each  angle  of  a  triangle  by  the  capital  letter  that  stands 
at  that  vertex;  and  to  use  the  corresponding  small  letter  to 
represent  the  length  of  the  side  opposite  that  vertex. 

Capital  letters  are  used,  then,  for  the  names  of  points 
and  for  angle-numbers;  small  letters  for  the  names  of  lines 
and  for  length-numbers.  Capital  letters  are  used  also  for 
area-numbers,  and  for  the  figures  having  the  areas  thus 
denoted. 

Thus,  in  the  triangle  ABC,  A 
repreaents  the  point  at  the  lower 
right-hand  comer,  a  the  aide  oppo- 
site A ;  A  represents  also  the  number  ^^^  5" 
of  degrees  in  the  angle  A,  and  a  the  p.  - 
number  of  feet,  inches,  or  other 
loigth-units  in  BC.    S  is  the  triangle  itself,  or  its  area-number. 

A  line  may  be  named  by  a  single  letter,  as  a;  or  by  two 
of  its  points,  as  BC. 

147 


148 


SOME  THEOREMS  OF  GEOMETRY 


An  angle  may  be  named  by  a  single  letter,  as  A;  or  by 
three  points  (one  from  each  side  and  the  vertex  in  the  middle), 
as  BAC. 

A  triangle  may  be  named  by  a  single  letter,  as  aS;  by  its 
three  vertices,  as  ABC;  or  by  its  three  sides,  as  abc. 

Some  Ratios  of  Area 

173.  Two  triangles  inscribed  in  the  same  stripe  have  the 
same  ratio  as  their  bases. 


Fig.  2. 


The  triangles  Si  and  S2  have  their  bases  61 
and  62  in  line,  and  their  vertices  are  at  one 
point.  Consequently  both  triangles  can  be 
inscribed  in  the  same  stripe. 

Let  p  represent  the  breadth  of  the  stripe. 

Then    ®    5i  =  i  hp  Formula 

0  same  values 

Equation  ©  is  the  algebraic  statement  of 
what  we  had  to  prove. 


0 

Si- 

=  hhip 

0 

S2 

=  hhp 

0 

!■ 

_hhp 
hh,p 

0 

Si 

S2" 

bi 

EXERCISE  61 

174.  Model  A.  —  Two  triangles  inscribed  in  the  same 
stripe  have  a  total  area  of  28.6  sq.  ft.,  and  their  bases  have 
a  ratio  1.04.     Find  the  area  of  each  triangle. 

Let     *Si  =  number  of  square  feet  in  area  of  first  triangle; 
^2  =  number  of  square  feet  in  area  of  second  triangle. 
Since  the  triangles  have  the  same  ratio  as  their  bases,  ^2  =  L04/Si. 
Whence  0     aSi  +  1.04  Si  =  28.6 

0  2.04  Si  =  28.6  0  same  values 

®  Si  =  14.0  0  -^  2.04 

®  1.04^1  =  14.6  0X1.04 

Ans.    The  areas  are  14.0  sq.  ft.  and  14.6  sq.  ft. 
Check:    14.0  +  14.6  =  28.6 


SOME  RATIOS  OF  AREA  149 

The  student  is  advised  to  draw  diagrams  for  the  problems  referring 
to  geometric  figures.    Even  a  rough  diagram  will  sometimes  help. 

Find  the  areas  of  the  following  triangles  which  are  in- 
scribed in  the  same  stripe: 

1.  Total  area  of  the  two  triangles,  79.5  sq.  ft.;  ratio  of 
bases,  3. 

2.  Total  area  of  the  two  triangles,  605  sq.  yd.;  ratio  of 
bases,  1.5. 

3.  Total  area  of  the  two  triangles,  100.6  sq.  in.;  ratio  of 
bases,  1.54. 

4.  Total  area  of  the  two  triangles,  91.1  sq.  cm.;  ratio  of 
bases,  2.11. 

5.  The  area  of  a  trapezoid  is  58.3  sq.  in.,,  and  the  two 
bases  have  a  ratio  of  1.39.  Find  the  areas  of  the  two  tri- 
angles formed  by  drawing  a  diagonal  of  the  trapezoid. 

6.  One  base  of  a  trapezoid  is  23  per  cent  less  than  the 
other;  and  the  area  of  the  trapezoid  is  5.63  sq.  ft.  Find  the 
areas  of  the  two  triangles  made  by  a  diagonal. 

If  one  base  is  23  per  cent  less  than  the  other,  it  is  .77  of  it;  that  is, 
the  ratio  of  the  bases  is  .77. 

175.  Model  B.  —  Of  two  triangles  inscribed  in  the 
same  stripe,  one  has  an  area  23  per  cent  greater  than  the 
other;  their  bases  differ  by  2.38  in.  The  breadth  of  the  stripe 
is  just  1  in.     Find  the  areas  of  the  triangles. 

Since  one  triangle  is  23  per  cent  greater  than  the  other,  the  ratio  of 
the  triangles  is  1.23;  but  since  they  are  in  the  same  stripe,  that  is  also 
the  ratio  of  their  bases;  wherefore 

Let  6i  =  number  of  inches  in  one  base. 

Then  1.23  6i  =  number  of  inches  in  other  base. 
©    1.23  6i  -  6i  =  2.38 

0  .23  &i  =  2.38  ®  same  values 

®  &i  =  10.35  ®  ^  .23 

0  1.23  6i  =  12.73  0  X  1.23 

.    Si  =  h  hxv  =  \  (10.35)  (1)  =  5  .18  sq.  in. 
Sa  =  \  (12.73)  (1)  =  6.36  sq.  in. 

An%.     The  areas  are  5.18  sq.  in.  and  6.36  sq.  in. 
Check:     .23  of  5.18  =  1.19;  and  5.18  +  1.19  =  6.37 


150  Some  theorems  of  geometry 

Find  the  areas  of  the  following  triangles  inscribed  in  the 
same  stripe: 

7.  Area  of  one  triangle  40  per  cent  more  than  the  other; 
difference  of  bases,  12  in.;  width  of  stripe,  40  in. 

8.  Area  of  one  triangle  53  per  cent  more  than  the  other; 
difference  of  bases,  74.2  ft. ;  width  of  stripe,  38.4  ft. 

9.  Area  of  one  triangle  53  per  cent  less  than  the  other; 
sum  of  bases,  46.7  yd.;  width  of  stripe,  21.6  yd. 

10.  Area  of  one  triangle  68  per  cent  more  than  the  other; 
sum  of  bases,  .783  rd.;  width  of  stripe,  .926  rd. 

11.  A  trapezoid  is  divided  by  its  diagonal  into  two  tri- 
angles, one  of  which  is  48  per  cent  greater  than  the  other. 
The  bases  of  the  trapezoid  differ  by  2.37  ft.,  and  the  width  of 
the  stripe  is  91.8  in.     Find  the  areas  of  the  triangles. 

176.  Model  C.  —  The  diagonal  of  a  trapezoid  2.36  in. 
high  divides  it  into  triangles  which  differ  by  .0394  sq.  ft. 
The  upper  base  is  23.6  per  cent  less  than  the  lower.  Find 
the  bases. 

Let  h  =  number  of  inches  in  lower  base. 

Then   .764  b  =  number  of  inches  in  upper  base.    (100%  —  23.6%) 
^  (2.36)  b  =  number  of  square  inches  in  area  of  larger  triangle. 
^  (2.36)  (.764  b)  =  number  of  square  inches  in  area  of  smaller  triangle. 


© 

.5  (2.36)  h 

1  -  .5  (2.36)  (.764  b) 

=  .0394  X  144 

© 

1.18  6  -  . 

,902  b  =  5.68 

©  same  values 

© 

.278  b  =  5.68 

©  same  values 

© 

b  =  20.4 

©  -^  .278 

© 

.764  b  =  15.6 

©  X  .764 

Ans.     The  bases 

are  20.4  and  15.6 

One  triangle, 

i  (2.36)  (20.4)  = 

24.1  sq.  in. 

Other  triangle 

;,    h  (2.36)  (15.6)  = 

18.4  sq.  in. 

Check: 


12.  The  diagonal  of  a  trapezoid  8  in.  high  divides  it  into 
triangles  which  differ  by  16  sq.  in.  The  upper  base  is  25 
per  cent  less  than  the  lower  base.     Find  the  bases. 


SOME  RATIOS  OF  AREA 


151 


13.  The  diagonal  of  a  trapezoid  .^3  in.  high  divides  it 
into  triangles  which  differ  by  .SM^sq.  in.  The  upper  base 
is  29.0  per  cent  more  than  the  lower  base.     Find  the  bases. 

14.  The  diagonal  of  a  trapezoid  3.00  ft.  high  divides  it 
into  triangles  which  differ  by  1.32  sq.  ft.  The  upper  base 
is  11. J  per  cent  less  than  the  lower.     Find  the  bases. 

177.  //  two  triangles  have  one  angle  in  each  the  same,  their 
areas  have  the  same  ratio  as  the  products  of  the  sides  that 
include  the  equal  angles. 

In  the  two  triangles  *Si  and  S2,  suppose  the  angle  made  by  the  lines  y 
and  2  is  A  degrees:    also  that 


the  angle  made  by  the  Unes  b 
and  c  is  A  degrees.  Then  a 
tracing  of  the  triangle  Si  can 
be  moved  over  S2  so  that  the 
angles  marked  A  will  exactly 
fit. 

Suppose  the  dotted  Une  in- 
dicates the  new  position  of 
the  base  of  Si.  Then  draw 
a  diagonal  in  the  figure 
{Si  —Si),  and  call  the  shaded 
triangle  X. 

The  triangles  Si  and  Si  -\-  X  can  be  inscribed  in  the  same  stripe. 
Then  their  ratio  is  equal  to  the  ratio  of  their  bases: 


Fig.  3. 


© 


Si 


Si+X 


The  triangles  Si  +  X  and  S2  can  also  be  inscribed  in  a  stripe. 


Then 


® 
® 


Si+X      z 


S2 


§1 
S2 


©  X0 


Therefore,  the  ratio  of  the  areas  of  the  triangles  Si  and  <S2  equals  the 
ratio  of  the  products,  yz  and  6c,  of  the  sides  that  include  the  equal 
angles. 


152 


SOME  THEOREMS  OF  GEOMETRY 


EXERCISE  62 
178.   Model  D.  —  The  side  AB  of  the  triangle  ABC  is 
28.3  ft.,  and  the  side  AC  is  36.2  ft.     On  AB  is  measured  off 
AX  =  13.6  ft.,  and  on  AC,  AY  =  11.8  ft.     What  fraction 
of  the  area  of  ABC  is  AXY?    What  per  cent? 

Since  ABC  and  AXY  have  A  as  a  common  angle,  their  areas  have  the 
ratio  described  in  Section  177: 


^    S2       be 

^     AXY       13.6  X  11.8 


® 
0 


ABC 
AXY 
ABC 
AXY 
ABC 


28.3  X  36.2 
161 
1025 

.157 


Formula 
0  given  values 
0  same  values 
0  same  values 


Check  : 


Ans.    AXY  is  .157  of  ABC,  or  15.7  per  cent. 
.157  (28.3  X  36.2)  =  13.6  X  11.8 
160.7  =  161 


1.  Triangles  ARS  and  FGH  have  angles  A  and  F  equal. 
AR  =  20  in.,  FG  =  35  in.,  AS  =  28  in.,  and  FH  =  60  in. 
What  fraction  of  FGH  is  the  triangle  ARS?    What  per  cent? 

2.  Triangles  HKL  and  HMN  have  angle  H  common. 
HK  =  13.2  cm.,  HL  =  21.3  cm.,  HM  =  12.3  cm.,  and 
HN  =  31.2  cm.  Find  what  part  of  HMN  is  the  triangle 
HKL.     Also,  what  per  cent  of  HMN  is  HKL? 

3.  Triangles  ABC  and  ADE  have  angle 
A  common.  AB  =  17.7  ft.,  AC  =  23.4  ft., 
AD  =  18.8  ft.,  and  AE  =  28.5  ft.  Find 
what  ratio  exists  between  the  areas  of  the 
triangles,  and  what  per  cent  each  one  is  of 
the  other. 

179.  Model  E.  —  In  the  triangle  XYZ, 
XY  is  28.7  ft.  long,  and  XZ  is  11.3  ft.  long. 
On  XF  the  point  P  is  marked,  and  on  XZ 
the  point  Q,  so  that  XP  =  XQ.    The  line  PQ 


Fig.  4. 


SOME  RATIOS  OF  AREA  153 

cuts  off  a  triangle  73  per  cent  less  than  the  area  of  XYZ. 
How  long  is  ZP? 

Let  X  =  number  of  feet  in  XP  and  in  XQ. 

Then,  since  XPQ  is  27  per  cent  of  XYZ  (100%  -  73%), 

^     XPQ        ^^  (x)  (x)  ^..      ._^ 

®      XYZ='^^  =  (28.7)  (11.3)  S««t^«^177 

®     .27  X  28.7  X  11.8  =  x^  ®  X  (28.7  X  11.3) 

0  87.5  =  x^  0  same  values 

0  ,  9.36  =  X  V0 

Ans.     XP  =  9.36  ft. 
Check:    9.36  X  9.36  =  (100%  -  73%)  28.7  X  11.3 
=  .27  (28.7  X  11.3) 
87.6  =  87.5 

4.  In  triangle  ADE,  AD  =  3.71  in.,  AE  =  4.38  in.,  and 
AB  along  AD  is  1^7  in.  C  is  a  point  on  AE.  Triangle  ABC 
is  38.1  per  cent  of  triangle  ADE.    How  long  is  AC? 

5.  In  triangle  MNO,  MN  =  4.50  in.,  MO  =  5.40  in.,  and 
MK  on  MO  is  twice  as  long  as  MH  on  MN.  Triangle  MHK 
is  44  per  cent  of  triangle  MNO.  Find  the  lengths  of  MH  and 
MK, 

6.  In  triangle  EFG,  EF  =  21.0  ft.,  EL 
on  EF  is  12.0  ft.,  while  EM  on  EG  is  15.6 
ft.  Triangle  ELM  is  55.8  per  cent  less 
than  triangle  EFG.     How  long  is  EG? 

180.  Model  F.  —  A  straight  Hne  RS  di- 
vides a  triangle  A  BC  into  a  smaller  triangle 
and  a  quadrilateral  (a  four-sided  figure),  Fig.  5. 

which  have  areas  of  16.2  sq.  in.  and  58.3 
sq.  in.,  respectively.      The  right  sides  of  the  two  triangles 
have  a  ratio  J  and  the  left  sides  differ  by  14.0  in.    Find 
the  lengths  of  the  left  sides. 


154  SOME  THEOREMS  OF  GEOMETRY 

Let  y  and  4  1/  =  number  of  inches  in  right  sides, 
X  and  a:  +  14  =  number  of  inches  in  left  sides. 


Check; 


(D 

ABC       16.2                a:i/ 

Section  177 

Ai^iS       74.5       {x  +  14)  4  2/ 

0 

16.2              X 

®  same  values 

74.5       {x  +  14)  4 

® 

4  (a;  +  14)  16.2  =  74.5  x 

@  X  4  (a:  +  14)  74.5 

® 

64.8  X  +  907.  =  74.5  x 

@  same  values 

® 

907.  =  9.7  X 

®- 64.8a; 

® 

93.8  =  a: 

®  -^9.7 

® 

107.8  =  x  +  14 

®  +  14 

Ans.     The  left  sides  { 

ire  93.8  in.  and  107.8  in. 

93.82/       _   93.8                         16.2 

=  .91 « 

107.8(4?/)       431.2       •  ^  74.5 


7.  A  straight  line  divides  a  triangle  into  a  smaller  triangle 
and  a  quadrilateral,  which  have  areas  of  32.0  sq.  ft.  and 
18.8  sq.  ft.,  respectively.  The  left  sides  of  the  two  triangles 
have  a  ratio  J,  and  the  right  sides  differ  by  5.10  ft.  Find 
the  lengths  of  the  right  sides. 

8.  A  straight  line  cuts  a  quadrilateral  whose  area  is  300 
sq.  ft.  from  a  triangle  whose  area  is  438  sq.  ft.  The  larger 
triangle  has  equal  sides  about  the  common  angle.  The 
smaller  triangle  has  one  side  \  of  one  of  the  equal  sides  of 
the  larger,  and  the  other  side  10  feet  less  than  one  of  these 
equal  sides.  Find  the  sides  of  the  triangles  about  the  com- 
mon angle. 

9.  A  quadrilateral,  whose  area  is  83.8  sq.  in.,  by  having 
two  of  its  sides  extended  forms  a  triangle  whose  area  is 
16.7  sq.  in.  One  side  of  the  triangle  is  equal  to  the  side  of  the 
quadrilateral  which  was  extended  to  form  it.  The  other  side 
of  the  triangle  is  17.3  in.  less  than  the  corresponding  side  of 
the  quadrilateral  extended  to  form  it.  Find  the  lengths  of 
the  two  lines  mentioned  in  the  last  sentence. 


-)    €/v"^ 


SIMILAR  FIGURES 
Similar  Figures 


155 


181.  A  polygon  is  a  flat  closed  figure,  the  boundary  of 
which  is  entirely  composed  of  straight  lines. 

If  two  polygons  contain  the  same  number  of  angles,  one 
angle  of  either  polygon  may  be  thought  of  as  corresponding 
to  an  angle  of  the  other  polygon,  the  next  angle  of  the  first 
polygon  to  the  next  angle  of  the  second,  and  so  on. 

Any  side  of  one  polygon,  lying  as  it  does  between  the 
vertices  of  two  angles,  may  then  be  said  to  correspond  to 
the  side  of  the  other  polygon  that  comes  between  the  corre- 
sponding angles. 

182.  Two  polygons  are  similar  when  the  corresponding 
angles  are  equal,  and  when  at  the  same  time  the  ratio  of  any 
two  corresponding  sides  is  equal  to  the  ratio  of  any  other  two 
corresponding  sides.  Except  in  the  case  of  triangles,  it  is 
possible  for  the  corresponding  angles  of  two  polygons  to  be 
equal  without  these  ratios  being  equal. 


Fig.  6. 

Thus,  in  the  two  polygons  K  and  M,  the  angles  that  are  lettered 
alike  are  equal.  These  angles  come  in  the  same  sequence  around  each 
polygon.     The  sides  a  and  x  come  between  corresponding  angles;  so  do 


156 


SOME  THEOREMS  OF  GEOMETRY 


b  and  y,  c  and  z.     Hence,  a  and  x  are  corresponding  sides;   b  and  y,  c 
and  z  also  correspond. 

The  ratio  -  is  greater  than  1;  -  is  less  than  1;  consequently,  -  5^  -• 
X  y  X      y 

Precisely  the  same  statements  are  true  of  the  rectangle  and  the 
square;   in  the  square,  of  course,  x  =  y. 

183.   The  following  theorem  about  triangles,  then,  is  not 
true  of  polygons  having  more  than  three  sides. 

//  two  triangles  have 
the  three  angles  of  one 
equal,  respectively,  to  the 
three  angles  of  the  other, 
the  ratio  of  any  two  corre- 
sponding sides  is  equal 
to  the  ratio  of  any  other 
two  corresponding  sides. 
That  is,  in  such  a  case, 
the  triangles  are  similar. 

In  the  two  triangles  ABC 
and  XYZ  (named  by  their 
vertices),  suppose  A  =  X,  B  =  Y,  C  =  Z;  represent  the  area  of  ABC 
by  Si  and  the  area  of  XYZ  by  S2. 

Si^bc 
S2  yz 
Si  _ca 
S2  ~  zx 
Si^ab 
S2  xy 
be      ca 


Fig.  7. 


Since  A  =  X, 
Since  B  =  Y, 
Since    C  =  Z, 


® 
® 
® 
© 
® 
® 
® 


yz      zx 
be      ab 


Section  177 
Section  177 
Section  177 
©  subst.  in  @ 


z      xy 
b_  a 
y~  X 
c  _a 
z~  X 


®  subst.  in  0 
c 


®- 


Hence,  the  ratios  -,  -,  and  -  are  all  equal. 
X    y  z 


SIMILAR  FIGURES  157 

EXERCISE  63 

1.  Is  the  following  statement  true?  If  two  right  triangles 
have  one  acute  angle  the  same,  they  are  similar.  Give  the 
reasons  for  your  answer. 

2.  Two  triangles  ABC  and  PQR  have  A  =  80°,  B  =  30°, 
Q  =  30°,  and  R  =  70°.  Are  the  triangles  similar?  State 
your  reasons. 

184.  Model  G.  —  Two  right  triangles  have  each  an  acute 
angle  of  73°.  The  longest  sides  differ  by  28.7  in.,  and  the 
shortest  sides  are,  respectively,  93.4  in.  and  80.7  in.  How 
long  are  the  longest  sides? 

Let  X  =  no.  of  inches  in  the  longest  side  of  one  triangle. 

Then    x  +  28.7  =  no.  of  inches  in  the  longest  side  of  the  other. 

®  ^^=9-M  ^^''-'^^ 

®  93.4  X  =  80.7  X  +  2314.  ©  X  93.4  {x  +  28.7) 

®  12.7  X  =  2316.  ®  -  80.7  x 

®  X  =  182.  ®  -J-  12.7 

0       x  +  28.7  =  211.  0  +  28.7 

Ans.     The  longest  sides  are  182  and  211  inches. 

80  7  182 

Check:     ^t-^  =  .864;  jr--  =  .860.     This  is  as  close  a  check  as  could 

be  expected,  considering  the  operation  in  0. 

-^  3.  Two  right  triangles  each  have  an  acute  angle  of  38°. 
Their  longest  sides  differ  by  30  in.,  and  their  shortest  sides 
are  50.0  in.  and  73.0  in.     Find  their  longest  sides. 

^  4.  Two  triangles  each  have  angles  of  68°  and  47°,  and 
their  shortest  sides  differ  by  18.3  in.  Their  longest  sides 
are  70.0  in.  and  43.4  in.     Find  the  shortest  sides. 

-  5.  One  right  triangle  has  an  acute  angle  of  49°,  another 
an  acute  angle  of  41°,  and  their  shortest  sides  are  40.3  in. 
and  61.9  in.,  respectively.  Find  their  longest  sides;  they 
differ  by  31.5  in. 


158 


SOME  THEOREMS  OF  GEOMETRY 


6.  One  right  triangle  has  an  acute  angle  of  37°,  another 
an  acute  angle  of  53°,  and  their  longest  sides  are  39.1  ft.  and 
54.7  ft.,  respectively.  Find  their  shortest  sides;  they  differ 
by  12.7  ft. 

v.  Two  triangles  are  similar;  the  largest  angle  of  one  is 
83.3°,  and  the  smallest  angle  of  the  other  is  41.8°.  What  are 
the  other  angles? 

185.  Model  H.  —  Two  triangles  are  similar;  two  sides  of 
one  are  128.3  ft.  and  101.7  ft.,  and  two  sides  of  the  other 

have  these  same  lengths.  In  the 
first  triangle  we  have  omitted 
the  shortest  side,  and  in  the 
second  the  longest.  What  are 
the  lengths  of  the  missing  sides? 

We  can  find  the  ratio  of  corre- 
sponding sides  in  these  two  triangles 
by  dividing  128.3  by  101.7  (the  sides 
of  medium  length).  This  gives  1.262 
for  the  ratio  of  any  side  in  the  larger 
triangle  to  its  corresponding  side  in 
the  smaller  triangle.      In  using  this 

ratio,  we  must  be  careful  always  to  have  for  the  numerator  one  of  the 

sides  of  the  larger  triangle. 


101.7 


Fig.  8. 


® 

(D 
® 
0 

® 


128.3 

t 

101.7 


=  1.262 
=  161.9 
=  1.262 


Section  183 
®  X  128.3 
Section  183 


101.7  =  1.262  s  ®Xs 

80.6  =  s  0  -^  1.262 

Ans.     The  missing  sides  are  80.6  ft.,  and  161.9  ft. 


Note  that  four-figure  accuracy  in  a  number  beginning  with  1  is  little 
better  than  three-figure  accuracy  in  a  number  beginning  with  8. 
161.9  _  101.7 
128.3       80.6 


Check: 


1.262 


SIMILAR  RIGHT  TRIANGLES  159 

In  the  following  table  of  values,  a,  6,  c,  and  a',  6',  c',  rep- 
resent the  corresponding  sides,  respectively,  of  two  similar 
triangles.     Find  the  lengths  of  the  missing  sides. 


a 

h 

c 

a' 

6' 

c' 

8. 

20 

30 

40 

50 

9. 

77 

85 

92 

69 

10. 

81.5 

77.4 

71.3 

64.2 

jLl. 

300 

360 

400 

460 

12. 

51.8 

69.7 

51.8 

69.7 

13. 

128.3 

201.7 

179.5 

162.4 

14.  Two  triangles,  ABC  and  PQR,  have  A  equal  to  72°, 
B  to  65°,  Q  to  65°,  and  R  to  43°.  Are  the  triangles  similar? 
State  your  reasons. 

15.  Two  triangles  are  similar;  the  largest  angle  of  one  is 
73.7°,  and  the  smallest  angle  of  the  other  is  45.5°.  What  are 
the  other  angles? 

Similar  Right  Triangles 

186.  If  we  know  one  acute  angle  of  a  right  triangle,  we 
know  that  the  ratios  of  the  sides  of  that  triangle  are  the 
same,  respectively,  as  the  ratios  of  every  right  triangle  that 
has  that  acute  angle. 

This  is  true  because  two  such  triangles  would  have  their 
corresponding  angles  equal.  The  two  right  angles  are  equal 
and  every  triangle  has  the  same  number  of  degrees  in  the 
sum  of  its  angles  (180°). 

For  example,  in  a  right  triangle  having  one  angle  equal  to  30°,  the 
ratio  of  the  two  perpendicular  sides  is  about  .577.  If  the  angle  is  20°, 
the  ratio  is  .364.  If  40°,  the  ratio  is  .839.  If  50°,  the  ratio  is  1.192.  If 
60°,  the  ratio  is  1.732. 

In  every  case,  the  ratio  given  is  the  ratio  of  the  side  opposite  the  acute 
angle  to  the  other  perpendicular  side. 


160  SOME  THEOREMS  OF  GEOMETRY 

187.  In  any  right  triangle,  the  ratio  of  the  side  opposite 
an  acute  angle  to  the  other  perpendicular  side  is  called  the 
tangent  of  that  angle. 

The  other  ratios  in  the  triangle  also  have  names,  but  for  the  present 
we  shall  consider  this  one  only. 

"  The  tangent  of  an  angle  of  23°  "  is  written  "  tan  23°." 

These  ratios  can  be  ascertained  by  accurately  construct- 
ing a  right  triangle  having  the  required  angle,  and  carefully 
measuring  the  sides.  There  are  other  and  more  accurate 
ways  of  ascertaining  them.  The  ratios  have  accordingly 
been  found  for  every  angle  that  one  is  likely  to  need,  and 
they  are  published  in  books  of  tables.  A  table  of  tangents 
for  every  degree  is  given  here. 


Table  of 

Tangents 

1° 

.0175 

16° 

.287 

31° 

.601 

46° 

1.036 

61° 

1.80 

76° 

4.01 

2° 

.0349 

17° 

.306 

32° 

.625 

47° 

1.072 

62° 

1.88 

77° 

4.33 

3° 

.0524 

18° 

.325 

33° 

.649 

48° 

1.11 

63° 

1.96 

78° 

4.70 

4** 

.0699 

19° 

.344 

34° 

.675 

49° 

1.15 

64° 

2.05 

79° 

5.14 

6° 

.0875 

20° 

.364 

35° 

.700 

50° 

1.19 

65° 

2.14 

80° 

5.67 

6° 

.105 

21° 

.384 

36° 

.727 

51° 

1.23 

66° 

2.25 

81° 

6.31 

7° 

.123 

22° 

.404 

37° 

.754 

52° 

1.28 

67° 

2.36 

82° 

7.11 

8° 

.141 

23° 

.424 

38° 

.781 

53° 

1.33 

68° 

2.48 

83° 

8.14 

9° 

.158 

24° 

.445 

39° 

.810 

54° 

1.38 

69° 

2.61 

84° 

9.51 

10" 

.176 

25° 

.466 

40° 

.839 

55° 

1.43 

70° 

2.75 

85° 

11.4 

ir 

.194 

26° 

.488 

41° 

.869 

56° 

1.48 

71° 

2.90 

86° 

14.3 

12° 

.213 

27° 

.510 

42° 

.900 

57° 

1.54 

72° 

3.08 

87° 

19.1 

13° 

.231 

28° 

.532 

43° 

.933 

58° 

1.60 

73° 

3.27 

88° 

28.6 

14° 

.249 

29° 

.554 

44° 

.966 

59° 

1.66 

74° 

3.49 

89° 

57.3 

16° 

.268 

30° 

.577 

45°  1 

.000 

60° 

1.73 

75° 

3.73 

EXERCISE  64 
•*•  188.   Model  I.  —  One  of  the  perpendicular  sides  of  a 
right  triangle  is  12.8  ft.,  and  the  acute  angle  next  to  it  is 
23°.    What  is  the  area  of  the  triangle? 


SIMILAR  RIGHT  TRIANGLES  161 

Let  X  =  the  number  of  feet  in  side  opposite  the  23°  angle. 

®     lis  ='^"23" 

(I)     -i_  =  .424        ®  tan  from  Table 
12. o 

®         X  =  5.43        0  X  12.8 

The  area  of  a  triangle  equals  |  base  j* 12.8  ft. H 

X  altitude;  hence,  the  area  of  this  tri-  F  r    Q 

angle  is  |  (5.43)  (12.8),  or  34.8  sq.  ft.  ^  '     * 


—   1.   One  leg  of  a  right  triangle  is  33.6  in.  and  the  angle 
opposite  is  28°.     What  is  the  area  of  the  triangle? 
The  perpendicular  sides  of  a  right  triangle  are  called  legs. 

^    2.   One  leg  of  a  right  triangle  is  74.9  ft.  and  the  adjacent 
angle  is  67°.     What  is  the  area  of  the  triangle? 

3.  The  two  legs  of  a  right  triangle  differ  by  3.87  ft.,  and 
the  acute  angle  next  to  the  larger  side  is  37°.  What  is  the 
length  of  each  of  these  sides? 

Let  X  and  a;  +  3.87  represent  the  lengths  of  the  sides,  and  form  an 
equation  by  means  of  tan  37°  taken  from  the  Table  of  Tangents. 

4.  The  two  legs  of  a  right  triangle  differ  by  43.8  in.,  and 
the  acute  angle  opposite  the  larger  leg  is  73°.  Find  the 
lengths  of  the  legs. 

5.  The  two  legs  of  a  right  triangle  have  a  sum  of  40  ft., 
and  one  acute  angle  of  the  triangle  is  26°.  Find  the  lengths 
of  the  legs  and  the  area  of  the  triangle. 

6.  A  building  lot  has  83.7  ft.  frontage,  and  the  angle  made 
with  the  front  line  by  the  diagonal  is  39°.  What  is  the 
perimeter? 

7.  A  rectangle  has  a  perimeter  of  71.8  in.,  and  the  diago- 
nal makes  with  one  edge  an  angle  of  73°.  Find  the  dimen- 
sions of  the  rectangle. 


162 


SOME  THEOREMS  OF  GEOMETRY 


8.  In  Fig.  10,  A  =  40°,  B  =  48°,  and  the  base  AB  =  7.39 
in.     Find  the  length  of  the  altitude  p. 

Let  X  =  the  number  of  inches  in  one  of  the  parts  of  the  base,  and 
7.39  —  X  the  number  of  inches  in  the  other.  Then  from  the  two  tangent 
equations  we  get  two  expressions  for  the  length  of  p,  which  give  us  an 
equation  for  x;  with  x  found,  we  can  obtain  p. 


Fig.  10. 

9.  In  Fig.  11,  A  =  27°,  B  =  43°,  and  AB  =  39.7  ft.    Find 
the  length  of  h. 

Use  a  method  similar  to  that  of  Ex.  8. 

10.  The  area  of  a  rectangle  is  4328  sq.  ft.  and  the  diagonal 
makes  with  one  side  an  angle  of  53°.     Find  the  dimensions. 

11.  The  area  of  a  right  triangle  is  32.84  sq.  in.,  and  one 
acute  angle  is  71°.     What  are  the  lengths  of  the  legs? 

Pythagorean  Theorem 

189.   The  famous  theorem  next  to  be  given  is  called  the 

Pythagorean  Theorem, 
after  Pythagoras,  a  pre- 
eminent Greek  geometer. 
It  was  at  least  partially 
known  centuries  before  his 
time. 

The  square  of  the  slant 
side  of  a  right  triangle  is 
equal  to  the   sum  of  the 
squares  of  the  perpendicular  sides. 


PYTHAGOREAN  THEOREM  163 

In  the  triangle  ahc,  the  line  a  is  perpendicular  to  b; 

therefore 

Again,  p  is  drawn  perpendicular  to  c; 
therefore 
and  also 

From  these  equations  we  obtain 

Since  these  three  right  triangles,  abc,  apx,  bpy,  have  their  acute  angles 
the  same,  they  are  similar.  Represent  the  area  of  apx  by  Si  and  that 
of  bpy  by  S2',  and  since        Si  '~  (aSi  +  S2) 


© 

A+B 

=  90 

0 

A  +  Y 

=  90 

® 

B  +X 

=  90 

® 

X 

=  A 

® 

Y 

=  B 

(D 

a  ~   c 

® 

y       b 
b        c 

.      [S2  ~  (5i  +  5i 

® 

ex  =  a2 

®  Xac 

® 

ey  =  ¥ 

®  X6c 

(S) 

ex  +  ey  =  a^ 

+  62 

®+® 

® 

c{x  +  y)  =a^ 

+  62 

@  same  values 

@ 

c2  =  a2 

+  62 

@[c  =  x  +  y] 

From  the  similar  triangles  Si  and  S2,  we  get  the  equations 

@  -  =  - 

^  p       y 

@  xy  =  p^  @Xpy 

Equations  ®,  @,  and  @  state  three  very  important  relations  foxmd 
in  every  right  triangle. 

The  sign  '^  means  "  is  similar  to." 

EXERCISE  65 

X  1-   The  two  perpendicular  sides  of  a  right  triangle  are 
85  ft.  and  132  ft.,  respectively.     How  long  is  the  slant  side? 

X  2.   Find  the  shortest  side  of  a  right  triangle  when  the 
other  two  sides  are  28.7  ft.  and  28.0  ft.,  respectively. 

3.  If  squares  are  constructed  on  the  three  sides  of  a  right 
triangle,  and  the  largest  one  is  three  times  the  area  of  the 
smallest  one,  what  is  the  ratio  of  the  squares  on  the  per- 
pendicular sides? 


164  SOME  THEOREMS  OF  GEOMETRY 

4.  The  slant  side  of  a  right  triangle  is  divided  into  two 
parts  by  the  altitude;  the  lengths  of  these  parts  are  54  ft. 
and  96  ft.,  respectively.     How  long  is  the  altitude? 

5.  The  distances  from  the  foot  of  the  altitude  of  a  right 
triangle  to  the  vertices  of  the  acute  angles  are  25  in.  and  12 
ft.,  respectively.     How  long  is  the  shortest  side? 

6.  The  altitude  on  the  slant  side  of  a  right  triangle  is 
23.6  ft.,  and  one  part  of  the  slant  side  from  the  foot  of  the 
altitude  to  a  vertex  is  15.9  ft.  Find  the  lengths  of  the  other 
lines  of  the  figure. 

>  7.  The  slant  side  of  a  right  triangle  is  34.2  in.,  and  the 
perpendicular  sides  are  in  the  ratio  2.31.  Find  the  perpen- 
dicular sides. 

\  8.  The  ratio  of  the  slant  side  of  a  right  triangle  to  one  of 
the  perpendicular  sides  is  3.55,  and  the  other  side  is  44.9  cm. 
Find  the  sides  of  the  triangle. 

9.  The  frame  of  a  kite  is  made  with  two  pieces  of  wood, 
30  in.  long  and  54  in.  long.  They  cross  at  right  angles  at 
the  middle  point  of  the  shorter  piece  and  18  in.  from  one  end 
of  the  longer.  Find  the  perimeter  of  the  kite,  and  its  area. 
^  10.  A  ladder  40  ft.  long  is  placed,  for  safety's  sake,  with 
its  foot  7  ft.  from  a  building.  How  far  up  the  side  of  the 
building  will  the  ladder  reach? 

Angles  of  a  Polygon 

190.  The  sum  of  the  angles  of  any  polygon  is  given  by 
the  formula  W  =  (n  —  2)  180°,  where  W  represents  the  sum 
of  the  angles  and  n  the  number  of  sides  of  the  polygon. 

Divide  the  polygon  into  triangles  by  drawing  all  possible  diagonals 
from  one  vertex. 

This  will  be  a  common  vertex  for  all  the  triangles.  Two  of  the  sides 
of  the  polygon  pass  through  this  common  vertex,  but  each  of  the  other 
sides  serves  as  the  base  for  one  triangle. 

There  will  be,  then,  just  n  —  2  triangles. 


ANGLES  OF  A  POLYGON 


165 


Letter  the  angles  of  each  triangle  with  the  same  letter  that  the  poly- 
gon had  at  that  point,  but  attach  suffixes  according  to  the  serial  number 
of  the  triangle;  that  is,  in  the  first  triangle  the  angles  would  be  Ai,  Bxy 
Ci)  in  the  second  A2,  C2,  Di]  in  the  third  A3,  D3,  Ez',  and  so  on. 


Then,  since  the  sum  of  the  angles  of  a  triangle  is  180°,  we  have  the 
following  series  of  equations: 

©  Ar  +  Bi  +  Ci  =  180 
®  A2  +  C2  +  D2  =  180 
®     A3  +  D3  +  ^3  =  180    and  so  on. 

There  would  be  n  —  2  of  these  equations,  one  for  each  triangle. 

Adding  the  equations,  we  should  find  Ai+A2  +  A3+  .  .  .  =A; 
Ci  +  C2  =  C;  D2  +  D3  =  D;  and  so  on.  In  the  last  two  triangles, 
(^6  +  (?6  =  G,  and  H^  =  H,  just  as  Bi  =  B. 

Then  the  sum  of  the  left  sides  of  all  the  equations  would  be  A  +  -B 
+  C  +  .  .  .  =  TF,  and  the  sum  of  the  right  sides  would  be  {n  —  2) 
180;  that  is,  W  =  {n  -  2)  180°. 

191.  An  equiangular  polygon  is  a  polygon  of  which  all  the 
angles  are  equal.  An  equilateral  polygon  is  one  of  which 
all  the  sides  are  equal. 

Except  in  the  case  of  a  triangle,  the  sides  of  a  polygon  do 
not  have  to  be  equal  when  the  angles  are  equal.  Thus  a 
square  and  a  rectangle  are  both  equiangular;  but  though  the 
sides  of  the  square  are  all  equal,  the  sides  of  the  rectangle 
are  not. 

192.  In  the  case  of  an  equiangular  triangle,  if  we  take  a 
tracing  of  the  triangle  and  turn  it  over  on  the  original  triangle 


166  SOME  THEOREMS  OF  GEOMETRY 

(making  the  tracing  of  one  side  fit  on  the  side  itself),  the 
angles  at  the  end  of  that  side  on  the  tracing  will  exactly  fit 
on  the  angles  of  the  triangle,  because  they  are  equal;  and  the 
other  two  sides  will  be  exactly  covered  by  their  tracings,  so 
that  the  other  vertex-angle  will  also  fit.    Hence  the  theorem: 

An  equiangular  triangle  is  also  equilateral. 

It  is  also  true  that  an  equilateral  triangle  is  equiangular. 

193.  A  regular  polygon  is  one  that  is  both  equilateral 
and  equiangular. 

Model   J.  —  How  many  sides  are  there  in  a  regular 
polygon  one  angle  of  which  is  175°? 
Let  n  =  number  of  sides  of  the  polygon. 

Then  175=^^?— ^H?5    and    n  =  72 

n 

EXERCISE  66 

1.  How  many  degrees  are  there  in  each  angle  of  an  equi- 
lateral triangle? 

2.  How  many  degrees  are  there  in  each  angle  of  a  regular 
5-sided  polygon? 

3.  How  many  degrees  in  each  angle  of  a  regular  polygon 
of  6  sides?  Of  10  sides?  Of  12  sides?  Of  180  sides?  Of 
100  sides?  Of  40  sides?  Of  n  sides?  Of  20  sides?  Of  23 
sides?     Of  33  sides? 

4.  How  many  sides  in  a  regular  polygon  one  angle  of 
which  is  144°? 

5.  How  many  sides  in  a  regular  polygon  one  angle  of 
which  is  174°?   176°?   172°?   156°?   170°?   150°?   168°? 


CHAPTER  IX 
THE  GRAPHICAL  METHOD 

194.  A  foot  rule  has  numbers  marked  for  each  inch,  and 
the  subdivisions,  though  not  marked,  are  known  to  cor- 
respond to  certain  definite  fractional  numbers.  In  the  same 
way,  any  straight  line  may  be  marked  at  convenient  equal 
distances,  starting  from  a  point  which  may  be  called  the 
zero  point.  Then  the  successive  marks,  or  their  distances 
from  the  zero  point,  represent  successive  whole  numbers; 
and  points  between  these  marks,  or  their  distances  from  the 
zero  point,  represent  fractional  numbers. 

195.  Some  points  on  such  a  line  cannot  be  expressed  in 
figures  because  they  represent  irrational  numbers.  Thus,  if 
we  construct  a  square  with  one  side  equal  to  the  (unit)  dis- 
tance between  two  successive  marks,  and  then  lay  off  on  the 
line  from  the  zero  point  a  distance  equal  to  the  diagonal  of 
that  square,  the  point  so  obtained  will  represent  \/2.  The 
number  tt  is  represented  by  a  point  which  can  be  found  by 
roUing  a  circle  with  unit  diameter  from  the  zero  point  one 
complete  turn  to  the  right. 

The  Algebraic  Scale 

196.  Points  in  regular  order  to  the  left  of  the  zero  point 
can  be  taken  to  represent  negative  numbers.  This  device 
has  the  advantage  that  the  distance  between  two  points  on 
the  line  represents  their  algebraic  difference.  If  the  distance 
runs  from  left  to  right,  the  difference  is  plus;  if  from  right 
to  left,  it  is  minus. 

167 


168  THE  GRAPHICAL  METHOD 

For  example,  in  Fig.  1,  the  points  marked  with  dots  are, 
respectively,  —6,  —2.5,  +3.5,  and  +8. 

Fig.  1. 

197.  A  Une  which  is  used  for  representing  numbers  in 
this  way  is  called  an  algebraic  scale. 

The  scale  of  a  thermometer  is  a  good  illustration  of  an 
algebraic  scale,  running  up  and  down  instead  of  right  and 
left.  The  rise  of  the  thermometer  is  the  algebraic  difference, 
going  up,  between  two  temperatures;  and  the  fall  is  the 
difference,  going  down,  —  a  negative  number. 


EXERCISE  67 

1.  Construct  an  algebraic  scale,  or  select  one  on  a  piece 
of  squared  paper,  and  on  it  mark  as  accurately  as  possible 
points  to  represent  -10,  -6,  -4.5,  0,  5,  8.5,  10.5,  \/2,  -  V2. 

2.  Draw  a  rectangle  2  units  long  and  one  unit  high. 
What  numbers  can  be  represented  by  marking  off  a  dis- 
tance, equal  to  its  diagonal,  to  the  right  and  to  the  left  of  the 
zero  point  on  an  algebraic  scale?  To  the  right  and  to  the 
left  of  the  point  representing  1?  What  approximate  deci- 
mal values  might  serve  to  locate  the  points  so  marked? 

3.  Draw  a  rectangle  one  unit  high,  of  which  the  base  ex- 
tends from  the  point  0  to  the  point  V2.  What  number  is 
represented  by  marking  off  a  distance  equal  to  its  diagonal? 

4.  On  an  algebraic  scale,  locate  the  point  —  \/6. 

5.  The  temperature  of  the  body  of  a  human  being  in  good 
health  is  98.4°  Fahrenheit;  the  temperature  at  which  copper 
melts  is  1900°  F.  These  two  points  cannot  be  accurately 
located  on  the  same  algebraic  scale.    Why  not? 


THE  ALGEBRAIC  SCALE 


169 


198.  When  two  numbers  are  related  to  each  other  so  that 
to  each  value  of  one  corresponds  one  value  of  the  other,  a 
diagram  can  be  made  in  which  each  pair  of  corresponding 
numbers  is  represented  by  one  point  on  the  diagram.  Ex- 
amples of  such  numbers  are  the  distances  traveled  by  a 
train  and  the  times  elapsed  since  starting;  the  premium 
charged  for  a  life  insurance  policy  of  $1000  and  the  age  of 
the  person  insured  at  the  date  of  the  insurance  contract;  the 
diameter  of  a  sphere  and  its  area. 


30 


10 


/ 

> 

/ 

/ 

y 

/ 

f/ 

/ 

,.<'■ 

■f 

/ 

w 

^^/ 

A 

f 

y 

7 

y 

/ 

/ 

V 

J 

/y 

/ 

1 

S    8 


Fig.  2. 


In  Fig.  2  is  shown  the  record  of  distance  and  time  for  two 
trains  from  Boston  to  Worcester,  a  total  distance  of  45 
miles.  This  diagram  is  constructed  with  hours  and  minutes 
along  the  bottom,  and  distances  up  the  side,  so  that  any  point 
on  the  diagram  represents  a  certain  distance  from  Boston 
and  a  certain  time  of  day.  Diagrams  of  this  kind  are  used 
in  train  dispatchers'  offices.  Any  point  on  the  line  marked 
/'Express  train''  shows  by  its  height  above  the  bottom  of 
the  diagram  how  far  the  train  has  gone;  and  by  its  distance 
from  the  left  side  of  the  diagram  it  shows  the  time  of  day. 
Similarly  for  every  point  on  the  line  marked  "Local  train." 


170 


THE  GRAPHICAL  METHOD 


When  the  express  train  overtakes  the  local,  the  intersection 
of  the  two  lines  warns  the  train  dispatcher  that  he  must 
provide,  at  the  place  and  time  indicated  by  the  point  of  in- 
tersection, opportunity  for  the  trains  to  pass. 

The  Algebraic  Diagram 

199.  Algebra,  however,  must  include  negative  numbers 
as  well  as  the  ordinary  numbers  of  arithmetic.  Conse- 
quently, an  algebraic  diagram  must  be  made  to  include 
representations  of  negative  numbers,  whether  paired  with 
each  other  or  paired  with  positive  numbers. 


X 


-4  -i  -2-1 


5     6     7 


Y 
Fig.  3. 

For  convenience  in  discussing  the  matter,  we  will  use  x 
and  y  to  represent  any  pair  of  corresponding  numbers. 
Our  diagram  is  constructed  by  taking  two  scale-lines  like  the 
one  described  in  Section  196,  and  placing  them  perpendicular 
to  each  other  with  the  zero  points  together. 

Call  the  horizontal  Hne  the  X-scale,  or  axis  of  Xy  and  the 
vertical  line  the  Y-scale,  or  axis  of  y. 

Then  imagine  vertical  hues  drawn  through  each  point  of 
the  axis  of  x,  and  horizontal  lines  drawn  through  each  point 
of  the  axis  of  y. 


THE  ALGEBRAIC  DIAGRAM 


171 


A  point  which  represents  x  =  —2,  y  =  4,  must  be  the 
point  which  Hes  opposite  —2  in  the  axis  of  x,  and  opposite  4 
in  the  axis  of  y. 


X 


Y 

n 

III 

1 

I 

I 

I 

1 

— 

_ 

5-< 

_• 

4-, 

0 

I  W  J 

i  i 

1 

o 

VI 

V 

• 

I 

V 

VI 

■ 

Y 

X 


Fig.  4. 

The  points  in  Fig.  4  that  represent  the  following  values  of  x  and  y  are 
marked  with  the  corresponding  Roman  numerals: 


I  X  =  4, 

II  X  =  -2 

III  X  =  3, 

IV  X  =  6, 

V  X  =  2i, 

VI  X  =  -2|,  y 

VII  X  =  -5,    ?/ 


2/  =  2 
^  =  3 
y  =  h 
2/=  -3 
2/=  -2^ 
-3 
-31 


The  construction  of  points  to  represent  in  this  way  cor- 
responding numbers  is  called  plotting. 

200.  The  size  of  the  space  between  successive  whole 
numbers  may  be  made  as  large  as  convenient,  and  with  large 
spaces  a  good  deal  of  subdivision  is  possible.  It  would  seem 
possible,  therefore,  to  secure  great  accuracy  in  a  diagram  by 


172  THE  GRAPHICAL  METHOD 

taking  a  large  unit  space.  When  this  is  done,  however,  the 
drawing  becomes  so  unwieldy  that  new  sources  of  error  arise 
and  the  results  lose  much  of  their  presumable  accuracy. 
With  small  diagrams  and  ordinary  care,  it  is  unhkely  that 
more  than  two  figures  can  be  properly  represented;  but  in 
large  and  careful  drawings  three  figures  can  be  used. 

When  data  are  given  more  accurately  than  the  diagram 
can  represent,  only  the  available  figures  are  used.  For  ex- 
ample, 38.07  and  57.93  would  probably  be  drawn  as  38 
and  58. 

201.  With  paper  already  ruled,  the  intervals,  or  spaces 
between  rulings,  may  be  made  to  correspond  to  units,  or  to 
hundredths,  or  to  tenths,  or  to  any  other  decimal  position 
of  the  digits. 

Thus,  in  plotting  temperatures  of  melting  for  metals,  the 
interval  might  well  represent  100°  Fahrenheit.  In  plotting 
the  temperatures  of  a  day,  the  interval  might  represent  1°; 
and  in  plotting  the  temperatures  of  a  fever  patient,  tenths  of 
a  degree  are  used.  The  "scale  of  the  diagram"  must  be 
decided  upon  in  each  case. 

EXERCISE  68 
V  1.  Insurance  companies  expect  a  person  in  good  health 
at  the  age  of  10  years  to  live  48.7  years  longer;  at  20  years, 
42.2  years  longer;  at  30  years,  35.3  years  longei-;  at  40  years, 
28.2  years  longer;  at  50  years,  20.9;  at  60  years,  14.1;  at 
70  years,  8.5;  at  80  years,  4.4;  at  90  years,  1.4.  Plot  the 
expectations  according  to  the  ages. 

2.  At  25  years  of  age,  one  must  pay  for  an  ordinary  life 
insurance  policy  $21.49  per  thousand;  at  30  years,  $24.38 
per  thousand;  at  35  years,  $28.11;  at  40  years,  $33.01;  at 
45  years,  $39.55;  at  50  years,  $48.48;  at  55  years,  $60.72; 
at  60  years,  $77.69;  at  65  years,  $101.48.  Plot  the  prices 
according  to  the  ages. 


THE  ALGEBRAIC  DIAGRAM  173 

y  8.  The  train  leaving  Boston  for  Portland  at  9  a.m.  is  12 
miles  away  at  9:21.  At  9:32,  it  is  17  miles  away;  at  10:05, 
40  miles;  at  10:35,  58  miles  (here  the  train  waits  ten  minutes) ; 
at  11:37,  100  miles;  at  12,  115  miles  (Portland). 

The  train  leaving  Portland  for  Boston  at  9  a.m.  is  100 
miles  from  Boston  at  9:26.  It  is  84  miles  away  at  9.44; 
70  miles  at  10:23;  58  miles  at  10:45  (here  the  train  waits 
ten  minutes);  40  miles  at  11:24;  17  miles  at  11:55;  12  miles 
at  12:06;  and  reaches  Boston  at  12:25. 

Plot,  on  a  single  diagram,  the  distances  from  Boston  corre- 
sponding to  the  times  for  each  train. 

y  4.  On  a  certain  Monday  the  moon  rises  at  8:35  p.m.  On 
Tuesday  it  rises  at  9:07;  Wednesday  at  9:34;  Thursday  at 
9:57;  Mday  at  10:25;  Saturday  at  10:51;  Sunday  at  11:21. 
Plot  the  times  of  rising  corresponding  to  the  days. 

5.  The  number  of  marketable  feet,  board  measure,  in  a 
spruce  tree  of  the  Adirondack  forests  is  64  if  the  diameter 
breast-high  is  9  inches,  83  if  the  diameter  is  10  inches, 
and  101  if  11  inches.  Then  127,  156,  189,  225,  263,  303, 
349,  396,  447,  503,  564,  630,  702,  776,  858,  948,  1013,  1089, 
1260  feet,  for  diameters  increasing  an  inch  at  a  time  up  to 
30  inches.  Compute  the  value  of  each  size  of  tree  at  $15 
per  thousand  feet,  and  plot  their  values  corresponding  to 
their  diameters. 

6.  The  Centigrade  and  the  Fahrenheit  scales,  for  ther- 
mometers, are  used  for  recording  the  same  temperatures, 
though  with  different  numbers.  Thus  32°  F.  is  the  same 
temperature  as  0°  C,  and  212°  F.  is  the  same  as  100°  C.  On 
a  diagram,  take  one  of  the  axes  to  represent  the  Centigrade 
scale,  the  other  the  Fahrenheit,  and  mark  on  the  diagram 
the  two  points  that  represent  the  temperatures  here  men- 
tioned. Now  suppose  that  all  points  representing  corre- 
sponding readings  of  the  two  scales  lie  in  line  with  the  two 


174  THE  GRAPHICAL  METHOD 

points  just  marked;  draw  the  line.  By  means  of  the  line 
you  have  drawn,  answer  the  following  questions: 

What  temperature  Centigrade  corresponds  to  68°  F.?  To 
104°  F.?    To  5°  F.?     To  -40°  F.? 

What  temperature  Fahrenheit  corresponds  to  15°  C?  To 
75°  C?    To-5°C.?    To-40°C? 

7.  The  centimeter  is  used  as  generally  on  the  continent 
of  Europe  as  the  inch  is  with  us,  and  much  scientific  work  in 
the  United  States  is  done  with  the  metric  system.  100 
centimeters  is  approximately  39.4  inches,  and  12  inches  is 
approximately  30.5  centimeters.  Take  one  axis  for  inches 
and  the  other  for  centimeters;  locate  the  two  points  de- 
scribed and  through  them  draw  a  straight  line.  By  means 
of  the  line,  answer  the  following  questions: 

How  many  centimeters  are  there  in  24  in.?  In  17  in.? 
In  50  in.?     In  1  in.?     In  44  in.?     In  29  in.? 

How  many  inches  are  there  in  20  cm.?  In  50  cm.?  In 
80  cm.?     In  16  cm.?     In  38  cm.?     In  53  cm.? 

202.  Model  A.  —  Plot  the  corresponding  numbers  which 
satisfy  the  equation  2x  —  Sy  =6. 

In  this  equation,  for  any  value  of  x  we  can  obtain  a  value  of  y.  Thus, 
if  a;  =  5,  2  a:  =  10,  and  we  have 


© 

10  - 

-3y  =  6 

® 

10  =  3  1/  +  6 

(D  +  ^y 

(D 

4  =  32/ 

@  -6 

0 

n  =  y 

®  -3 

Again,  if  a;  =  6,  ?/  =  2;  if  x  =  —3,  y  =  —4;  and  so  on. 
Thus  we  obtain  the  following  table  of  corresponding  numbers: 

X  =  -  ^,  y  =  -  ^  X  =  1,  y  =  -  Ih  X  =  5,  y  =  H 

x=-S,  y=-4:  X  =  2,  y  =  -I  x  =  Q,  y  =  2 

X  =  -2,  y  =  -Sk  X  =  S,  y  =  0  x  =  1,  y  =  2% 

X  =  0,  y  =  -2  a;  =  4,  2/  =  |  x  =  8,  y  =  3i 


THE  LOCUS  OF  AN  EQUATION 


175 


These  pairs  of  numbers  will  be  represented  by  the  points  in  Fig.  5, 
running  up  toward  the  right. 


Fig.  5. 

The  Locus  of  an  Equation 

203.  It  would  be  found,  on  trial,  that  all  the  other  pairs 
of  corresponding  numbers  which  satisfy  this  equation  lie  on 
the  same  line,  i.e.,  their  points  fall  into  line  with  those 
already  put  down.  In  fact,  the  straight  line  drawn  through 
two  points  that  represent  any  two  answers  to  the  equation 
is  a  complete  list  of  all  possible  pairs  of  corresponding  num- 
bers that  satisfy  the  equation.  In  order  to  be  complete, 
this  line  and  the  two  axes  are  supposed  to  be  endless  in  extent. 

This  Hne  is  called  the  locus  of  such  points  as  satisfy  the 
equation,  or  for  brevity,  the  locus  of  the  equation.  The 
word  locus  is  the  Latin  word  for  place;  it  means  here  the 
place  to  find  all  such  points,  and  no  others. 

204.  If  for  any  temperature  /  on  the  Fahrenheit  scale  the 
reading  on  the  Centigrade  scale  is  c,  the  two  readings  are 
connected  by  the  equation 

©/  =  |c  +  32 
This  is  an  explicit  formula  for  /. 


176 


THE  GRAPHICAL  METHOD 


In  the  same  way  the  equation 

@c  =  |(/-32) 

is  an  explicit  formula  for  c. 
And  the  equation 

®  5/-9C  =  160 

is  an  implicit  formula  for  either  /  or  c. 

Show  how  equations  ®  or  0  could  be  derived  from  ®. 

Define  an  explicit  formula.     Also  an  implicit  formula. 

Notice  that  an  answer  to  a  two-letter  equation  is  a  pair  of  numbers 
that  will  satisfy  the  equation.  We  are  said  to  plot  the  equation  when 
we  plot  such  pairs  of  numbers. 

Plot  the  equations  ®,  ®,  and  ®.  Why  are  they  all  represented  by 
the  same  Une? 

205.  With  the  information  we  already  have  about  similar 
triangles,  it  is  easy  to  prove  that  every  straight  line  in  an 
algebraic  diagram  is  the  locus  of  an  equation  of  the  first  degree. 

In  the  accompanying  figure, 
where  a  straight  line  crosses  the 
axis  OX  at  A,  and  the  axis  07  at 
B,  OA  and  OB  are  constant;  let 
us  represent  them  by  a  and  6, 
respectively. 

Then,    if    P   represents    any 
point  on  the  line  AB,  OM  and 
MP  are  the  variables  that  would 
be  represented  by  x  and  y,  re- 
spectively.    In  other  words,  a 
and  h  are  two  numbers  repre- 
sented by  the  points  at,  which  this 
line  crosses  the  axes;  x  and  y  are  the  two  numbers  represented  by  any 
point  on  the  line  AB;  they  will  vary  as  we  take  different  points  on  the 
line. 

For  any  one  position  of  the  point  P,  such  as  the  one  shown  in  the 
illustration,  there  are  two  right  triangles,  which  are  similar. 

AMP  ~  AOB.     (Under  what    condition   are   two   right    triangles 
similar?     How  do  these  two  right  triangles  fulfill  this  condition?) 
In  these  two  triangles,  MP  {=  y)  in  the  triangle  AMP  corresponds  to 


Y 

B 

^V^ 

^ 

\. 

"^ 

^       X 

0 

1 

r~ 

'^ 

^\ 

Fig.  6. 


THE  LOCUS  OF  AN  EQUATION  177 

DB  {=  b)  in  the  triangle  OAB;  and  MA  (=  a  —  x)   corresponds  to 
OA  (=  a).     Then  we  have  the  equation 

®  ^ — L 

^  a      a  —  X 

If  we  multiply  each  side  of  this  equation  by  a,  we  get 

6=  "y 


a  —  X 

Then,  multiplying  by  (a  —  a;),  h  (a  —  x)  =  ay 

Or,  in  one  operation  ®     b  {a  —  x)  =  ay  (i)  X  a  (a  —  x) 

This  equation  can  be  changed  in  form  thus 

®       ab  —  bx  =  ay  0  same  values 

0  ab  =  bx  +  ay  (3)   +  bx 

Thus  we  have  established  the  fact  that  every  point  on  the 
straight  Une  represents  two  numbers  that  satisfy  a  certain 
equation  of  the  first  degree,  the  coefficients  of  which  can  be 
ascertained  in  advance  by  the  measurements  of  the  figure. 

Take  a  point  not  on  the  straight  line,  and  call  it  Q.  Draw  QN  ±  OX, 
cutting  AB  as  at  R.  Then  R  represents  a  number-pair  that  satisfies 
our  equation.  Does  Q?  Can  you  give  an  algebraic  reason  for  your 
answer?     Draw  your  own  diagram. 

The  sign  ±  means  "perpendicular  to." 

EXERCISE  69 

1.  Draw  the  diagram  and  follow  through  the  proof  given 
in  Section  205  for  the  case  where  OA  =  9  and  OB  =  4. 

2.  Proceed  as  in  Ex.  1,  for  the  case  where  OA  =  —  3  and 
OB  =  5;  only,  for  the  sake  of  simplicity,  mark  the  point  P 
on  the  right  of  OY. 

3.  In  the  same  way,  find  the  equation  of  the  line  AB 
where  a  =  —4,  and  6  =  1. 

4.  Find  also  the  equation  of  the  line  AB  where  a  =  2, 
b  =  -5. 

5.  In  the  equation  Sx  —  4y  =36,  what  value  of  x  cor- 
responds to  the  value  2/  =  0?    Where  is  the  point  that  rep- 


178  THE  GRAPHICAL  METHOD 

resents  this  pair  of  values?  What  value  of  y  corresponds  to 
the  value  a:  =  0?  Where  is  the  point  that  represents  this 
pair  of  values? 

206.  The  points  where  a  line  crosses  the  axes,  in  an 
algebraic  diagram,  represent  numbers  that  are  called  the 
intercepts  of  the  equation.  Thus,  in  the  demonstration  of 
Section  205,  a  is  the  intercept  on  OX;  b  is  the  intercept  on 
OY, 

The  intercepts  of  an  equation,  then,  may  be  defined  as  the 
points  that  satisfy  the  equation  and  lie  on  the  axes. 

The  intercept  on  OX  will  have  y  =  0.  The  value  of  x 
corresponding  to  y  =  0  can  be  found  from  the  equation,  and 
will  be  the  value  of  a.  In  the  same  way,  the  value  of  b  can 
be  found  by  letting  x  =  0  in  the  equation. 

EXERCISE  70 

1.  What  are  the  intercepts  of  the  equation  2  x  -\-  5  y  =  40? 
Follow  through  the  proof  of  Section  205  for  the  equation  of 
a  straight  hne  having  these  intercepts. 

2.  What  are  the  intercepts  of  the  equation  Sy  —  4:X  =  60? 
Follow  through  the  proof,  in  the  same  way,  for  the  equation 
of  the  straight  fine  having  these  intercepts. 

3.  Proceed  in  the  same  way  for  the  equation  3y  -\-  9  =  2x. 

4.  Find  the  intercepts  of  the  equation  S  x  -{-  7  y  =27. 
What  is  the  tangent  of  the  angle  OAB  in  the  algebraic  dia- 
gram for  this  equation? 

5.  The  point  2,  3  satisfies  the  equation  in  Ex.  4.  Mark 
the  point  N  on  OX  to  represent  x  =  2;  draw  NQ  J.  OX, 
making  NQ  =  3.  What  is  the  tangent  of  NAQ'!  How 
does  this  show  whether  Q  is  on  AB? 

207.  From  the  preceding  examples  it  is  clear  that  for  any 
two-letter  equation  of  the  first  degree  we  can  find  a  pair  of 


THE  LOCUS  OF  AN  EQUATION        179 

intercepts  and  a  straight  line  passing  through  them.  With- 
out a  formal  proof,  for  the  present,  we  may  take  it  for  granted 
that  the  equation  we  start  with  will  be  the  equation  of  the 
straight  line  so  obtained. 

We  may  conclude,  then,  that  every  two-letter  equation  of 
the  first  degree  has  for  its  locus  a  straight  line. 

It  requires  some  reflection  to  see  that  this  statement 
differs  from  that  of  Section  205;  but  it  is  very  important  to 
appreciate  the  difference.  It  becomes  clearer  when  the 
statements  are  expressed  in  the  conditional  form,  thus: 

In  the  case  of  an  equation  for  which  we  have  plotted  the 
locus  on  an  algebraic  diagram, 

(1)  If  the  locus  is  a  straight  line,  its  equation  is  a  two-letter 
equation  of  the, first  degree. 

(2)  //  the  equation  is  a  two-letter  equation  of  the  first  degree^ 
its  locus  is  a  straight  line. 

Two  statements  related  in  this  way,  so  that  the  con- 
ditional part  of  each  is  the  conclusion  of  the  other,  are 
called  converse  statements. 

/ 

EXERCISE  71 

For  each  of  the  following  statements,  state  the  converse. 
Then  answer,  for  each,  these  two  questions:  first.  Is  the 
original  statement  true?     Second,  Is  its  converse  true? 

1.  If  a  man  is  very  rich,  he  can  afford  to  pay  his  carfare. 

2.  If  Leander  has  broken  his  leg,  he  cannot  swim  two 
miles. 

3.  If  it  is  raining,  there  are  clouds  in  the  sky. 

4.  If  I  am  your  teacher,  you  are  my  pupil. 

5.  If  his  name  is  on  the  voting  list,  he  can  vote. 

6.  If  a  man  plays  the  cornet  badly,  he  annoys  his 
neighbors. 


180  THE  GRAPHICAL  METHOD 

7.  If  you  feel  tired,  you  have  walked  ten  miles  to-day. 

8.  If  my  straw  hat  is  ruined;  a  horse  has  stepped  on  it. 

9.  If  Caius  lived  in  Caesar's  time,  Caius  has  been  dead 
for  centuries. 

10.  If  this  book  were  written  in  Greek,  it  would  be  hard 
to  understand. 


CHAPTER  X 
ELIMINATION  BY  COMBINATION 

208.  An  equation  with  two  unknown  letters  imposes  a 
condition  upon  the  values  of  those  letters.  For  example, 
the  equation  2  x  —  S  y  =  Q  is  an  impUcit  formula  for  y 
when  the  value  of  x  is  given,  or  for  x  when  the  value  of  y  is 
given.  Thus  it  imposes  a  condition  upon  the  two  numbers 
x  and  y,  such  that  when  a  numerical  value  is  assigned  to 
either  there  is  one  and  only  one  corresponding  value  of 
the  other.  The  nature  of  the  condition  imposed  by  the 
equation  is  indicated  by  drawing  its  locus.  Every  point  of 
that  locus  represents  a  pair  of  numbers  which  will  satisfy 
the  equation. 

Two  equations,  with  two  unknown  letters  in  each,  may 
be  plotted  upon  the  same  algebraic  diagram.  If  the  condi- 
tions expressed  by  these  equations  are  understood  to  be 
imposed  upon  the  same  pair  of  numbers,  —  in  other  words; 
if  the  abbreviations  x  and  y  have  the  same  meaning  in  both 
equations, — then  the  equations  are  called  simultaneous. 
The  conditions  that  they  impose  can  be  satisfied  by  the 
points  that  He  upon  both  loci  and  by  those  only. 

To  determine  the  values  of  two  unknown  letters  two  conditions 
are  necessary. 

EXERCISE  72 

Construct  the  loci  of  the  following  equations,  putting  each 
pair  on  a  separate  diagram: 

1.  x-{-2y  =5,         Sx  +  7y  =  17 

2.  15  X  +  3  2/  =  39,  17  a;  -  2/  =  31 

181 


182  ELIMINATION  BY  COMBINATION 

3.  7x  +  4?/=41,  Saj  — 2/=4 

4.  2x  +  2/  =  11,  3 re +  7 2/  =22 

5.  5x  +  62/=51,  6a:-52/  =  -12 
e.  2x-  y  =  -4,  7x  +  Qy  =Q2 

It  will  be  found  that  these  pairs  of  lines  intersect  in  the 
points  given  below;  the  points  are  numbered  to  correspond, 
with  the  pairs  of  equations.  Each  of  these  points,  then, 
satisfies  both  equations  in  the  pair  to  which  it  corresponds. 

1.  X  =  1,  y  =  2      3.  X  =  S,  y  =  5       5,  x  =  S,  y  =  Q 

2.  X  =  2,  y  =  'S      4.   x=5,  2/  =  l        e.   X  =  2,  y  =  S 

Two-letter  Problems 

209.  Model  A.  — Six  horses  and  11  cows  sell  for  $810; 
13  horses  and  5  cows  sell  at  the  same  rate  for  $1190.  Find 
the  price  of  each  animal. 

Let    X  =  the  number  of  dollars  for  one  horse; 
y  =  the  number  of  dollars  for  one  cow. 

(I)  Qx  +  ny  =  SlO 

(II)  13  a;  +    6y  =  1190 

210.  Model  B.  — Six  pounds  of  tea  and  11  pounds  of 
coffee  sell  for  $9.23;  13  pounds  of  tea  and  5  pounds  of  coffee 
sell  at  the  same  rate  for  $11.90.     What  is  the  price  of  each? 

Let    X  =  the  number  of  cents  for  one  pound  of  tea; 
y  =  the  number  of  cents  for  one  pound  of  coffee. 

(III)  '  Qx  +  Uy  =  923 

(IV)  13  a;  +    5y  =  1190 

211.  In  the  four  equations  that  come  from  these  two 
problems,  x  and  y  do  not  have  the  same  meaning.  In  (I) 
and  (II),  for  example,  x  stands  for  the  number  of  dollars  in 
the  price  of  a  horse;  in  (III)  and  (IV),  for  the  number  of  cents 
in  the  price  of  a  pound  of  tea.  When  the  unknown  letters 
in  two  equations  stand  for  entirely  different  quantities,  as  in 


TWO-LETTER  PROBLEMS  183 

(II)  and  (III),  they  must  be  treated  as  if  they  were  written 
with  different  letters;  the  terms  which  seem  to  be  similar 
are  really  not  similar.  In  (I)  and  (II),  we  should  find  that 
X  =  $80  and  y  =  $30;  while  in  (III)  and  (IV)  we  should 
find  that  x  =  75  cents  and  ?/  =  43  cents. 

212.  In  two  simultaneous  equations,  however,  terms  hav- 
ing the  same  letters  are  similar  terms.  Any  member  of 
either  equation  can  be  added  to  or  subtracted  from  a  member 
of  the  other  equation,  and  if  the  x  terms  (or  the  y  terms)  are 
the  same,  a  resulting  equation  may  be  obtained  free  from  x 
(or  y).    Thus  for  Model  A: 

(I)  6rc  +  ll2/  =  810 

(II)  13a:+    5y  =  1190 

®        30  x  +  55  2/  =  4050  (I)    X  5 

0       143  X  +  55  2/  =  13090  (II)  X  11 

The  y  terms  are  now  alike.  Subtracting  ®  from  0, 

0  113  a;  =  9040  0-0 

0  a;  =  80  0  -^  113 

To  find  the  value  of  y,  we  may  return  to  the  original 
equations,  (I)  and  (II).  We  make  the  x  terms  alike  by 
multiplying  (I)  by  13  and  (II)  by  6,  and  then  subtract  the 
resulting  equations.  Or,  since  we  know  now  that  x  =  80, 
Q  X  =  480,  we  may  write  instead  of  (I), 

0         480  +  11 2/  =  810  0  subst.  in  (I) 

.0  11 2/  =  330  0  -  480 

0  2/  =  30  0  -^  11 

The  answer  to  Model  A,  then,  is  $80  for  each  horse,  and 
$30  for  each  cow.  In  the  same  way  we  find,  for  Model  B, 
75  cents  a  pound  for  tea,  and  43  cents  a  pound  for  coffee. 

This  is  the  method  by  which  the  points  of  crossing  were 
found  for  the  six  pairs  of  lines  in  Exercise  72.  If  the  x 
terms  (or  the  y  terms)  are  of  opposite  signs  in  the  two  equa- 
tions, the  equations  must  be  added,  instead  of  subtracted,  to 


184 


ELIMINATION  BY  COMBINATION 


get  rid  of  that  letter.  The  algebraic  work  really  effects  the 
reduction  of  the  two  given  equations  to  two  new  equations 
in  the  form  of  particular  values  for  x  and  y. 

Model  A.     Check: 

6  horses  at  $80  =  $480 

11  cows  at  $30  =  330 

•     $810 


13  horses  at  $80  =  $1040 

5  cows     at  $30  =      150 

$1190 


Model  B.     Check: 

6  pounds  at  .75  =  $4.50 

11  pounds  at  .43  =    4.73 

$9.23 


13  pounds  at  .75  =  $9.75 

5  pounds  at  .  43  =    2.15 

$11.90 


EXERCISE  73 

Solve  the  following  simultaneous  equations  for  x  and  y^ 
choosing  for  elimination  the  letter  which  requires  less  work : 

1.  2x  +  2/=5,     3x  +  2/=7 

2.  2x-\-y  =  n,     x  +  2y=7 

3.  3x  +  22/  =  13,     x  +  2y  =7 

4.  3  X  +  4 !/  =  13,     4  a;  +  2/  =  13 

6.  x-\-y  =  17,    x-y  =  5 

e.  x  +  dy  =7,    Sx  +  5y  =  13 

7.  2x  +  5y  =  17,    4:X  +  Sy  =  13 
s.  x  +  2y  =  12,    x-2y  =  -S 
9.  2x  +  y  =  12,    Sx  +  2y  =  19 

10.  x-\-10y  =  IS,  x  +  Qy  =9 

11.  2 re  +  5 2/  =  29,     Sx-2y  =  -4: 

12.  3x  +  2/  =  ll,  10x-Sy=5 

13.  5x  +  42/=27,  5y  —  4:X  =  —5 

14.  Qx  +  5y  =27,  5x-\-y  =  1S 

15.  Qx  —  dy  =  9,  5x  —  y  =  17 
le.  2x-  y  =  -2,  Sx  +  y  =7 


\ 


TWO-LETTER  PROBLEMS 


185 


17. 

18. 

19. 

20. 

21. 

22. 

23. 

24. 
^  25. 

26. 

27. 

28. 

29. 
^30. 

31. 

32. 

33. 

34. 

35. 

36. 

37. 

38. 

39. 
<3r40. 

41. 

42. 

43. 

44. 

45. 


2x  +  y  =  lh     dx-y=9 
2x-y  =2,    dx-  y  =  5 
3x  +  4:y=S2,    4x-2/  =  ll 
Sx  —  4:y  =  —l,    4:X  —  y  =  lQ 
2x-\-Sy  =27,     3x-2y  =S 
x  +  y  =  11,    x-Sy  =  -13 
Sx-  5y  =  -27,     5x-Sy  =27 
4:X  +  5y=SS,     X  —  y  =  —4: 
3x  +  42/=22,     2x-y  =  ll 
x  +  Sy  =  12,     Sx-2y  =  14: 
2x  +  52/=36,     Sx-4y  =  -15 
3x  +  42/=36,     2a:-52/  =  -22 
6 X  -  2/  =  20,    Qx  +  5y  =80 
5x+10y  =  15,     5x-Sy  =2 
2x-y  =9,     Sx+lOy  =4:8 
2x  +  y  =  16,     3x-\-10y  =  58 
5x-4y  =4Q,    4 a;  +  10  ?/  =  50 
lOx  +  Sy  =40,     5x-2y  =  -15 
-2y  =  -18 


4x 


Sy  =  -22,  a; 
3x  +  52/=35,  5a;-22/=48 
5x-Sy  =44,  4x-  lOy  =2\ 
Qx  +  5y  =41,  5x-  4y  =2Q 
Qx-  5y  =2Q,  5x  +  4y  =S8 
6x  —  5y  =  17,    5x  —  4?/  =  15 


Qx  —  5y  =  17,  5x  —  4?/  =  15 

5x  +  3y  =  53,  4x-  lOy  =  -32 

5x-32/=5,  4a:+10  2/=66 

5  X  +  3  2/  =  46,  3  X  +  10  2/  =  85 

5a;-32/=25,  3x-10  2/  =  -26 

5  a;  +  3  2/  =  38,  3  a;  -  10  2/  =  11 


^ 


186  ELIMINATION  BY  COMBINATION 

46.  5  rc  -  3  2/  =  37,     3  x  +  10  2/  =  34 

47.  3x  -  51/  =  -41,     5 a; +  2  2/  =35 

48.  3a;  -  52/  =  -38,     5a;-22/=0 

49.  3x  —  42/  =  18,     2a;  —  52/=5 

50.  3  X  -  4  2/  =  14,     2  a;  +  5  2/  =  40 

213.  The  process  of  getting  rid  of  an  unknown  letter  is 
called  eliminating  that  letter.  The  method  used  above 
is  called  elimination  by  combination.  There  is  another 
method,  which  will  be  described  later. 

214.  The  student  is  advised  to  check  his  numerical  work 
in  advance  by  constructing  the  loci  of  the  equations  he  is 
to  work  upon  and  finding  the  point  where  they  intersect. 
The  most  convenient  way  of  constructing  the  loci  is  by 
means  of  the  intercepts.  Instead  of  actually  drawing  the 
lines  on  the  algebraic  diagram,  the  student  will  find  it  easier 
to  take  a  sheet  of  tracing  paper  on  which  a  straight  line  has 
been  drawn.  Lay  the  tracing  paper  over  the  algebraic 
diagram  so  that  the  line  lies  on  the  intercepts  of  one  of  the 
two  equations.  Lay  a  ruler  with  its  edge  on  the  other  two 
intercepts;  and  with  a  pin  mark  the  point  where  this  edge 
crosses  the  tracing-paper  line.  This  point  should  correspond 
to  the  values  of  x  and  y  in  the  answer. 

EXERCISE  74 
Solve  the  following  equations: 

1.  2a;  +  72/=36,     5x  +  32/=61 

2.  5a;  +  82/  =34,     9a;  +  22/=24 

3.  13  X  +  17  2/  =  107,     2  X  +  2/  =  10 

4.  14x  +  92/  =  100,     7x  +  22/=30 
6.  X  +  15  2/  =  37,     3  X  +  7  2/  =  35 

6.   15  X  +  19  2/  =  79,    35  X  +  17  2/  =  157 


TWO-LETTER  PROBLEMS  187 

7.  6  a;  +  4  2/  =  90,     3  x  +  15  2/  =  123 

8.  39  a;  +  27 1/  =  213,     52y  +  29x=  318 
q9.   18  a; +  98  2/  =664,     Q3x  +  7y  =  112 

10.  26  a;  +  101 2/  =  511,     103  a;  +  39  2/  =  941 

"^11.  2  x  +  7  ^  =  49,     6  a;  -  5  2/  =  17 

12.  7  a;  -  4  2/  =  2,     25  2/  -  13  X  =  49 

j^    13.  a;  +  2/  =  90,     x  —  2/  =  13 

14.  4  x  +  9  2/  =  50,     7  x  -  17  2/  =  22 

^       15.  5  a;  -  3  2/  =  17,     12  2/  -  7  a;  =  23 

r    16.  171  x-21Sy  =  642,     114  a;  -  326  2/  =  244 

17.  9  2/  -  7  a;  =  43,     15  X  -  7  2/  =  43 

18.  12  a;  +  7  2/  =  176,     3  2/  -  19  x  =  3 

19.  43  x  +  2  2/  =  266,     12  a:  -  17  2/  =  4 
020.  5x4-92/  =  188,     13 X  -  2 2/  =  57 

21.  4  X  +  3  2/  =  22,     3  X  +  5  2/  =  11 

22.  5  X  -  4  2/  =  57,     17  X  -  3  2/  =  162 

23.  3x  —  52/  =  51,     5x  +  72/=39 

24.  7  2/  -  5  X  =  143,     3  X  +  5  2/  =89 
Q)25.  4x+ll2/  =  144,     172/  — 8x=414 

26.  2x-72/=8,     32/-9x=21 

27.  17  X  +  12  2/  =  59,     19  X  -  3  2/  =  148 

28.  8x  +  32/=3,     4x  +  32/  =  l 
C29.  59  2/  -  17  X  =  123,     2x  -  13  2/  =  -17 

30.   3X  +  22/  =42,     13x  +  232/=225 

EXERCISE   75. 

1.  For  8  cows  I  get  $20  more  than  I  pay  for  40  sheep; 
and  for  16  sheep  I  pay  $21  more  than  I  receive  for  3  cows. 
Find  the  price  of  each  animal. 


188  ELIMINATION  BY  COMBINATION 

'  2.   Find  two  numbers  whose  difference  is  3V  of  their  sum, 
and  is  3  less  than  ^  of  the  larger  number. 

3.  In  10  hours  A  walks  1  mile  less  than  B  does  in  8  hours; 
and  in  6  hours  B  walks  1  mile  less  than  A  does  in  8  hours. 
How  many  miles  does  each  walk  per  hour? 

4.  If  A's  money  were  increased  by  36  cents,  he  would 
have  three  times  as  much  as  B;  if  B's  money  were  increased 
by  5  cents,  he  would  have  half  as  much  as  A.  How  much 
money  has  each? 

6.  A  pound  of  tea  and  6  pounds  of  sugar  cost  72  cents; 
if  sugar  were  to  rise  50  per  cent  and  tea  10  per  cent,  the  same 
quantity  would  cost  84  cents.  Find  the  price  of  the  tea 
and  of  the  sugar. 

6.  Find  two  numbers  such  that  3  times  the  greater  ex- 
ceeds twice  the  less  by  29,  and  twice  the  greater  exceeds 
3  times  the  less  by  1. 

7.  Three  men  and  16  boys  together  earn  $107.25  in  5§ 
days,  and  4  men  with  10  boys  earn  $192.50  in  11  days. 
What  are  the  daily  wages  of  each  of  these  men  and  boys? 

8.  A  farmer  bought  land  for  $7500,  part  of  it  at  $80  per 
acre  and  part  at  $50  per  acre.  The  cheaper  land  he  sold  at 
a  loss  of  10  per  cent,  and  the  more  valuable  land  at  a  gain  of 
10  per  cent.  On  the  whole,  he  profited  $50  by  the  trans- 
action.    How  much  land  did  he  buy? 

9.  If  I  divide  the  smaller  of  two  numbers  by  the  greater, 
the  quotient  is  .36  and  the  remainder  .64.  If  the  greater  of 
the  two  numbers  is  divided  by  the  smaller,  the  quotient  is  2 
and  the  remainder  27.     Find  the  numbers. 

-i- 10.  The  sum  of  the  ages  of  a  father  and  a  son  will  be 
doubled  in  25  years;  the  difference  of  their  ages  is  J  of  what 
the  sum  will  be  in  20  years.    How  old  are  they? 


TWO-LETTER  PROBLEMS  189 

11.  Find  two  numbers  such  that  3  times  the  greater  ex- 
ceeds I  the  less  by  439,  and  3  times  the  less  exceeds  |  the 
greater  by  49. 

12.  Eight  years  ago  A  was  5  times  as  old  as  B;  7  years 
hence  he  will  be  only  twice  as  old.  How  old  are  they 
now? 

--  13.  A  merchant  offers  me  8  pounds  of  black  tea  and  24 
pounds  of  green  tea  for  $10,  or  14  pounds  of  the  black  tea 
and  17  pounds  of  the  green  tea  for  the  same  sum.  If  it 
makes  no  difference  to  him  which  offer  I  accept,  what  are 
his  prices  per  pound? 

14.  A  and  B  buy  a  horse  for  $550.  A  can  pay  for  it  if  B 
will  advance  f  of  the  money  he  has  in  his  pocket;  and  B 
can  pay  for  it  if  A  will  advance  J  of  the  money  in  his  pocket. 
How  much  money  has  each  man? 

15.  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons.  If  there  had  been  10  more  persons, 
each  would  have  received  $2  less;  if  there  had  been  10  less, 
each  would  have  received  $3  more.  How  many  persons 
were  there  and  what  was  the  share  of  each? 

16.  If  a  rectangular  lot  of  land  were  8  feet  wider  and  2 
feet  longer,  it  would  contain  960  square  feet  more.  If  it 
were  2  feet  narrower  and  8  feet  shorter,  it  would  contain  760 
square  feet  less.     What  is  its  area? 

17.  A  farmer  bought  a  certain  quantity  of  eggs  at  2  for  5 
cents,  and  other  eggs  at  3  for  8  cents.  Wishing  to  sell  them 
to  a  relatiye  at  cost,  he  named  a  price  of  31  cents  per  dozen, 
but  found  he  would  lose  5  cents.  His  relative  then  sug- 
gested 5  for  13  cents,  and  the  farmer  accepted  that  price, 
but  made  a  profit  of  10  cents  on  the  transaction.  How  many 
dozen  were  there  of  each  kind? 


190  ELIMINATION  BY  COMBINATION 

18.  It  takes  me  16  times  as  long  to  walk  around  the  edge  of 
a  long  rectangular  field  as  to  walk  directly  across  it.  The 
next  field,  which  is  30  feet  wider  and  150  feet  shorter,  has 
the  same  area.     What  are  the  dimensions  of  each  field? 

19.  An  electric  car  40  feet  long,  running  parallel  to  a  rail- 
road, passes  a  train  455  feet  long,  running  in  the  opposite 
direction,  in  9  seconds.  The  electric  car  is  later  overtaken 
and  passed  in  15  seconds  by  the  same  train  returning  at  the 
same  speed.  How  fast  was  the  car  going?  How  fast  was 
the  train  going? 

20.  A  cruiser  386.8  feet  long  passes  a  battleship  going  in 
the  same  direction  in  1.12  minutes;  returning,  she  passes 
the  same  battleship  in  12  seconds.  The  length  of  the  battle- 
ship is  352.4  feet.  Supposing  each  ship  maintains  the  same 
speed  throughout,  how  many  miles  per  hour  do  they  go? 
(5280  feet  =  1  mile.) 

Find  the  fractions  described  as  follows: 

21.  If  4  is  added  to  the  numerator,  the  value  of  the 
fraction  will  become  1.  If  3  is  added  to  the  denominator, 
the  value  of  the  fraction  will  become  |. 

22.  If  1  is  subtracted  from  the  numerator,  the  value  of  the 
fraction  will  become  J.  If  11  is  added  to  the  denominator, 
the  value  of  the  fraction  will  become  J. 

23.  If  8  is  added  to  the  numerator,  the  value  of  the 
fraction  will  become  f .  If  8  is  added  to  the  denominator, 
the  value  of  the  fraction  will  become  J. 

24.  If  17  is  added  to  numerator  and  tO  denominator,  the 
value  of  the  fraction  will  become  |.  If  10  is  subtracted 
from  numerator  and  from  denominator,  the  value  of  the 
fraction  will  become  |. 

25.  If  3  is  taken  from  the  numerator  and  1  is  added  to 
the  denominator,  the  value  of  the  fraction  will  become  |. 


TWO-LETTER  PROBLEMS  191 

If  3  is  added  to  the  numerator  and  5  to  the  denominator, 
the  value  of  the  fraction  will  become  |. 

0  26.  Find  two  fractions  with  numerators  4  and  11  re- 
spectively, such  that  their  sum  is  2Jf ,  and  if  their  denomina- 
tors are  interchanged  their  sum  is  2|J. 

27.  A  fraction  is  such  that,  when  it  is  multiplied  by  f ,  the 
sum  of  its  terms  is  107.  If  8  is  added  to  each  of  its  terms, 
the  value  of  the  fraction  becomes  |. 

28.  A  proper  fraction  is  such  that  if  the  numerator  is 
halved  and  the  denominator  increased  by  3,  its  value  be- 
comes i.  If  the  fraction  is  multiplied  by  |,  the  difference  of 
its  terms  will  be  53. 

29.  Two  fractions  have  denominators  20  and  10,  re- 
spectively. The  fraction  formed  from  these  two  by  taking 
for  its  respective  terms  the  sums  of  the  corresponding  terms 
of  these  two  fractions  is  f ;  and  the  fraction  similarly  formed 
by  taking  the  differences  is  f . 

Fiiid  the  numbers  described  as  follows.  Unless  otherwise 
stated,  the  number  has  two  digits. 

30.  A  number  divided  by  the  sum  of  its  digits  gives  7f 
for  a  quotient.  If  54  is  subtracted  from  the  number,  the 
digits  are  interchanged. 

31.  If  I  divide  a  number  by  the  sum  of  its  digits,  the 
quotient  is  6  and  the  remainder  7.  But  if  I  invert  the  order 
of  the  digits  and  then  divide  by  the  sum  of  the  digits,  the 
quotient  is  4  and  the  remainder  6. 

32.  A  number  exceeds  5  times  the  sum  of  its  digits  by  8. 
If  the  order  of  the  digits  were  reversed,  the  number  would 
be  15  less  than  7  times  the  sum  of  the  digits. 

33.  A  number  is  4  times  the  sum  of  its  digits;  and  if  27 
is  added  to  the  number,  the  order  of  the  digits  is  interchanged. 


192  ELIMINATION  BY  COMBINATION 

34.  A  number  is  two  more  than  5  times  the  difference  of 
its  digits.  If  its  digits  are  interchanged,  the  resulting  num- 
ber is  8  times  the  sum  of  its  digits. 

35.  Two  numbers  have  the  same  digits  in  opposite  order. 
The  difference  of  the  numbers  is  3  times  the  sum  of  the 
digits;  and  the  sum  of  the  digits  is  2  more  than  2|  times  the 
difference  of  the  digits. 

36.  If  45  is  subtracted  from  a  number,  the  digits  are  in- 
terchanged; but  the  same  result  might  have  been  obtained 
by  subtracting  7  from  the  number  and  then  dividing  by  2. 

37.  Two  numbers  which  have  the  same  digits  in  opposite 
order  differ  by  18;  and  the  smaller  number  is  4  times  the 
sum  of  the  digits. 

38.  If  a  certain  number  is  divided  by  the  sum  of  its 
digits,  the  quotient  is  4  and  the  remainder  6.  If  the  digits 
are  interchanged,  the  resulting  number  divided  by  5  gives  4 
more  than  the  sum  of  the  digits. 

39.  A  number  consists  of  3  digits,  the  middle  digit  being 
zero.  If  the  order  of  the  digits  is  reversed,  the  number  is 
increased  by  396.  If  the  second  and  the  third  digits  change 
places,  the  number  is  increased  by  3  more  than  6  times  the 
sum  of  the  digits. 

40.  A  number  consists  of  three  digits,  the  middle  digit 
being  6.  It  has  the  order  of  its  digits  reversed  if  we  add  to 
it  22  times  the  sum  of  its  digits;  and  the  number  so  formed 
is  72  less  than  double  the  original  number. 

41.  There  is  a  number  between  1000  and  10,000,  of  which 
the  second  digit  is  6  and  the  fourth  is  3.  Reversing  the 
order  of  the  digits  increases  the  number  by  909;  but  if  only 
the  first  and  third  digits  change  places,  the  number  is  in- 
creased by  2970. 


TWO-LETTER  PROBLEMS  193 

The  following  numbers  may  be  used  to  form  new  problems 
for  Exs.  1-20.  Substitute  the  numbers  here,  in  the  order 
given,  for  the  numbers  in  the  corresponding  problem. 

That  is,  the  problem  formed  from  Ex.  1  of  Set  A  will  read:  For  5  cows 
I  get  $45  more  than  I  pay  for  20  sheep;  and  for  19  sheep  I  pay  $48  more 
than  I  receive  for  2  cows. 

Set  a.  1.  5,  45,  20,  19,  48,  2.  11.  5,  i  508,  5,  |,  404. 

2.  },d,l  12.  3,3,7,2. 

3.  6,  2,  4,  4,  1,  7.  13.  20,  12,  15,  10,  21. 

4.  9,  4,  4,  I  14.  100,  i  i 

5.  10,  8,  5.88,  10,  25,  6.54.  15.  6,  8,  6,  12. 

6.  5,  3,  1.06,  3,  5,  22.  16.  6,  10,  1260,  5,  10,  1000. 

7.  5,  3,  80.50,  7,  2,  7,  18.75,  2^ 

8.  1860,  75,  60,  5,  20,  57.  17.  4,  3,  3,  4,  12,  4,  9,  10,  44. 

9.  .28,  .76,  3,  3.  18.  12,  12,  48. 
10.  60,  i,  20.  19.  38,  298,  7,  14. 

20.   380,  lj\,  13,  335. 

Set  B. 


1. 

8,  80,  15,  20,  35,  5. 

11. 

4,  i  302,  4,  i,  268. 

2. 

A,  2,  h 

12. 

4,  4,  10,  2. 

3. 

8,  1,  6,  4,  1,  6. 

13. 

60,  25,  52,  64,  20. 

4. 

18,  5,  21,  i 

14. 

340,  i,  I 

5. 

3,  5,  1.96,  25,  5,  2.37. 

15. 

8,  2,  8,  3. 

6. 

7,  4,  1.77,  4,  7,  54. 

16. 

5,  8,  1600,  5,  2,  1100. 

7. 

8,  5,  92,  4,  3,  1,  38.75 

,5. 

17. 

4,  7,  6,  13,  25,  10,  20,  39,  6. 

8. 

2760,  88,  72,  12^  15, 

18. 

18. 

7,  42,  60. 

9. 

.47,  .03,  2,  3. 

19. 

30,  245,  5,  11. 

10. 

33,  h  41. 

20. 

383.5,  1.4,  12,  355.7. 

Set  C.  1.  4,  109,  11,  20,  165,  3.  11.  2,  |,  31,  2,  i,  36. 

2.  h  14,  i  12.  6,  2,  9,  U. 

3.  12,  3,  10,  8,  -1,  10.  13.  24,  40,  44,  11,  55. 

4.  20,  5,  2,  i.  14.  430,  I  |. 

5.  2,  10,  1.50,  20,  30,  1.87.  15.  6,  5,  6,  7. 

6.  11,  8,  1.84,  8,  11,  25.  16.  7,  2,  776,  3,  8,  824. 

7.  2,  12,  204,  Sh  5,  7,  76.50,  3.  17.  3,  2,  6,  5,  9,  1,  14,  11,  20. 

8.  3100,  49,  38,  25,  20,  107.  18.  15,  50,  200. 

9.  .33,  .66,  2,  32.  19.  37,  243,  7,  20. 

10.  36,  i  4.  20.  211.2,  1^,  11,  369,  6. 


194  ELIMINATION  BY  COMBINATION 

Set  D.  1.  8,  50,  26,  13,  30,  3.  11.  6,  i  282,  6,  |,  208. 

2.  ^Q,  4,  }.  12.  8,  7,  3,  3. 

3.  15,  -5,  10,  8,  -4,  9.  13.  12,  20,  9,  8,  15. 

4.  21,  6,  14,  i  14.  620,  i  ^V- 

5.  2,  8,  1.30,  30,  20,  1.65.  15.  6,  5,  6,  15. 

6.  19,  16,  200,  16,  19,  -25.  16.  4,  7,  672,  6,  3,  528. 

7.  7,  5,  128,  8,  4,  11,  106.75,  7.  17.  6,  7,  3,  5,  17,  3,  2,  3,  68. 

8.  3360,  35,  28,  5,  7h,  77.  18.  8,  27,  54. 

9.  .78,  .22,  1,  23.  19.  48,  502,  11,  25. 

10.  43,  i  26.  20.  391.8,  1.5,  14,  301.2. 

Set  E.  1.  7,  105,  5,  18,  26,  5.  11.  7,  },  483,  7,  },  333. 

2.  h  3,  h  12.  7,  10,  6,  4. 

3.  10,  -3,  7,  18,  5,  15.  13.  20,  20,  14,  18,  23. 

4.  28,  7,  1,  i  14.  365,  ^\,  ^J^. 

5.  5,  1,  2.28,  5,  12i,  2.40.  15.  25,  9,  25,  15. 

6.  17,  13,  1060,  13,  17,  -460.  16.  9,  3,  543,  1,  7,  257. 

7.  3,  5,  68.75,  5,  10,  13,  103.125,  21. 

8.  10000,  90,  70,  7,  9,  452.  17.  6,  11,  3,  7,  25,  3,  6,  13,  48. 

9.  .56,  .4,  1,  15.  18.  10,  77,  154. 
10.  63,  },  63.  19.  37,  443,  8,  20. 

20.  398.1,  1.45,  15,  367.5. 

Solving  for  Reciprocals 
215.   In  equations  like  the  following  it  is  better  to  make 
the  X  terms  (or  the  y  terms)  alike  without  first  clearing  of 

fractions.    The  reciprocal  of  x  is  -. 

X 


Model   C.  —  Solve  the  f 

oUowing  pair 

equations: 

^     3,2      2 

^5,4      11 
®    i+-y  =  2i 

®  hH 

©  X2 

®     l-ii 

©-© 

©          21  =  x 

©  X  21  2; 

The  value  of  y  may  be  found  in  a  similar  way. 


wb>^ 


I    ..  SOLVING  FOR  RECIPROCALS  195 

EXERCISE  76 

1.  A  and  B  together  can  do  a  piece  of  work  in  8f  days. 
If  A  worked  3  days  and  B  5  days,  only  half  of  the  work 
would  be  done.  How  long  would  it  take  A  alone  to  do  the 
work?     How  long  for  B  alone? 

2.  Five  boys  and  10  men  could  do  a  certain  job  in  3  days; 
one  man  and  one  boy  would  take  24  days  to  do  it.  How 
long  would  it  take  one  man  alone  to  do  the  work?  One  boy 
alone? 

3.  Twenty-four  pails  and  20  cans  of  water  will  just  fill  a 
certain  tank;  6  pails  and  14  cans  will  half  fill  it.  How  many 
pailfuls  will  fill  it?     How  many  canfuls? 

4.  Two  pipes  fill  a  cistern  in  20  minutes;  if  the  first  pipe 
were  twice  as  large  and  the  other  half  as  large,  the  cistern 
would  be  filled  in  15  minutes.  How  long  would  it  take  to 
fill  the  cistern,  using  only  the  first  pipe?  The  second? 
(The  word  large  here  refers  not  to  the  width  but  to  the 
capacity  of  the  pipe.) 

5.  Two  pipes  can  fill  a  cistern  in  5  minutes;  if  one  of  the 
pipes  is  closed  half  the  time,  it  will  take  7|  minutes.  How 
long  would  it  take,  using  each  pipe  alone? 

6.  A  and  B  together  could  dig  a  well  in  12  days;  but  at 
the  end  of  the  third  day  B  quits,  so  that  the  job  takes  A 
6  days  longer  than  it  otherwise  would.  How  long  would 
it  take  each  man  working  alone? 

7.  After  working  6  days  on  a  certain  job  with  B,  A  says 
to  him,  "I  can  finish  this  job  alone  in  10  days."  B  replies, 
''If  we  work  together  one  more  day,  I  can  finish  it  alone  in  5 
days."  If  what  they  say  is  true,  how  long  will  it  take  each 
alone  to  finish  the  job? 

8.  A  can  row  11  miles  downstream  in  the  time  it  takes  him 
to  row  7  miles  against  the  stream.     He  rows  downstream  for 


196  ELIMINATION  BY  COMBINATION 

3  hours,  then  rows  back,  and  at  the  end  of  6  hours  he  is  5 
miles  from  his  starting  place.  How  fast  does  he  row? 
How  fast  does  the  stream  flow? 

9.  B  rows  9  miles  downstream  in  45  minutes.  He  rows 
back,  near  the  bank  where  the  current  is  only  half  strength, 
in  one  hour  and  a  half.  Find  the  speed  of  the  boat  and  of 
the  stream. 

10.  C  rows  6  hours  downstream  and  15  hours  upstream, 
covering  in  all  72  miles.  His  speed  upstream  is  f  of  his 
speed  in  still  water.     What  is  the  speed  of  the  stream? 

11.  D  rows  downstream  for  an  hour  and  a  half,  but  it  takes 
him  3  hours  to  row  back.  In  the  first  three  hours  he  rows 
12  miles.     What  is  the  speed  of  the  boat  and  of  the  stream? 

12.  A  and  B  run  a  mile  race.  A  gives  B  12  seconds  start 
and  beats  him  by  44  yards.  Then  A  gives  B  165  yards  start, 
and  is  beaten  by  10  seconds.     What  is  the  speed  of  each? 

13.  A  traveler  started  on  a  journey  of  330  miles,  having 

4  hours  and  12  minutes  to  spare  to  make  connections  with 
another  train  at  the  end  of  that  journey.  An  accident, 
which  occurred  when  he  was  2  hours  out,  held  up  the  train 
for  2  hours,  and  diminished  its  speed  for  the  rest  of  the  run. 
Then  he  missed  his  connection  by  just  three  minutes.  If 
the  accident  had  happened  6  miles  farther  on,  he  would 
have  been  just  in  time  to  make  the  connection.  How  fast 
did  the  train  go  before  the  accident?  How  fast  did  it  go 
after  the  accident? 

14.  A  was  sent  to  a  town  147  miles  away,  and  7  hours 
later  B  was  sent  after  him.  After  traveling  71  miles,  B  was 
handed  a  letter  to  deliver  to  a  person  living  17  miles  beyond 
the  town  to  which  A  had  been  sent.  He  overtook  A  just  as 
he  was  entering  the  town,  and  handed  him  the  letter;  the 
letter  was  delivered  just  9  hours  and  40  minutes  after  B  re- 
ceived it.    At  what  speed  did  A  travel?    B? 


WHERE  ELIMINATION  FAILS  197 

Where  Elimination  Fails 

216.  Model  D.  —  A  certain  number  consisting  of  two 
digits  is  equal  to  4  times  the  sum  of  its  digits.  If  the  digits 
are  interchanged,  it  is  equal  to  21  times  the  difference  of  its 
digits.     Find  the  number. 

Let  X  =  the  tens  digit; 

y  =  the  units  digit. 
Then         x  -\- y  —  the  sum  of  the  digits. 

X  —  y  =  the  difference  of  the  digits. 
I®  X  -\-  y  =  the  number. 

lOy  -{■  X  =  the  number  with  the  digits  interchanged. 
©         lOx  +  y  ^4{x  +  y) 
(D         10y  +  x  =  21ix-y) 


®  10x-\-y  =  4iX-\-4Ly  ©  same  values 

0  Qx-dy=0  ®  -  4.x  -  4:y 

©  2x-y  =  0  ©  -^3 

®  lOy  -\-  X  =  21x  —  21y  @  same  values 

®3l2/-20a;  =  0  ®-21x  +  21y 

©  20  a;  -  10  y  =  0  ®  X  10 

®  2ly  =  0  ©+© 

@  y  =  0  ®  -T-  21 

@  2  a;  =  0  @  subst.  in  ® 

@  a;  =  0  @  ^  2 

This  result  is  absurd,  since  a  number  with  zero  digits  is 
no  number  at  all.  Such  a  result  points  to  some  mistake  in 
the  statement  of  the  problem,  or  in  our  understanding  of  it. 

217.  We  took  for  granted,  in  forming  equation  ®,  that 
the  tens  digit  was  larger  than  the  units.  Let  us  see  if  that 
is  the  source  of  the  absurd  result. 

©  10x  +  ?/  =  4x  +  4i/ 

@  lOy +  x  =  21y -2lx 


®22a;-ll2/  =  0  ®  +  21x-21|/ 

@  2x-y  =  0  ®-Ml 

Equation  ®  is  the  same  as  ©. 


198  ELIMINATION  BY  COMBINATION 

In  this  case  also,  then,  we  cannot  determine,  from  the 
two  equations  given,  the  particular  values  of  x  and  2/, 
because  the  two  conditions  are  not  independent. 

218.  Model  E.  —  A  certain  number  of  two  digits,  if 
added  to  half  the  sum  of  the  digits,  gives  a  total  of  21.  The 
units  digit  added  to  7  times  the  tens  digit  gives  a  total  of 
26.     Find  the  number. 


Let                X  =  the  tens  digit; 

y  =  the  units  digit. 

Then     a;  +  ?/  =  the  sum  of  the  digits. 

10  .-c  +  7/  =  the  number. 

®         10  X  +  2/  4-  H^  +  2/)  =  21 

(D                                7  a;  +  2/  =  26 

®           20x  +  2i/  +  a;  +  2/  =  42 

®  X2 

®                           21  X  +  3  y  =  42 

®  same  values 

®                                 7  a:  +  2/  =  14 

0-3 

Equations  (2)  and  ©  cannot  possibly  be  true  for  the  same 
values  of  x  and  y.  Such  equations  are  called  inconsistent. 
Draw  their  loci  and  find  out  how  this  inconsistency  is  repre- 
sented on  the  algebraic  diagram. 

Non-Algebraic  Conditions 

219.  It  happens  sometimes  that  a  problem  is  given,  like 
Model  D,  which  implies  only  one  algebraic  condition,  but 
still  the  answer  is  restricted  to  a  few  particular  values,  or 
even  to  only  one.  Such  a  restriction  is  due  to  a  condition, 
implied  in  the  statement  of  the  problem,  such  that  it  cannot 
be  stated  algebraically. 

In  Model  D,  our  digits  must  be  from  the  nine  Arabic 
numerals.  We  found  by  trial  that  our  result  was  absurd 
unless  the  second  digit  were  the  larger;  and  our  algebraic 
conditions  both  reduced  to  the  fact  that  the  second  digit 


NON-ALGEBRAIC  CONDITIONS  199 

was  twice  the  first.    That  leaves,  for  the  only  possible 

solutions: 

X  =  1, 2/  =  2;  the  number  12 

X  =  2,  2/  =  4;  the  number  24 

x  =  3, 2/  =  6;  the  number  36 

X  =  4,  y  =  S;  the  number  48 

Four  particular  answers  to  the  problem. 

Check:  12  =  4  (1  +  2);  21  =  21  (2  -  1);  and  similarly  for  the 
other  three  numbers. 

220.  Equations  which  do  not  give  a  solution  from  the 
algebraic  conditions  alone  are  called  indeterminate  equa- 
tions. The  problems  which  give  rise  to  such  equations, 
although  they  may  contain  enough  non-algebraic  con- 
ditions to  determine  the  answer,  are  often  called  inde- 
terminate problems. 

Find  three  numbers,  each  of  which  will  satisfy  one  of  the  two  in- 
consistent conditions  in  Model  E. 


CHAPTER  XI 
ELIMINATION  BY  SUBSTITUTION 

221.  A  two-letter  equation  of  the  first  degree  is  called  a 
linear  equation. 

Any  equation  in  which  at  least  one  term  contains  two 
unknown  letters  as  factors,  and  no  term  contains  more 
than  two,  is  called  a  quadratic  equation. 

These  two  unknown  letters  that  are  thus  used  as  factors 
of  a  term  may  be  the  same  letter,  as  in  the  terms  5  x^  and 
32  2/^;  or  they  may  be  different  letters,  as  in  the  term  3  xy. 
The  equations  that  follow  are  examples  of  two-letter  quad- 
ratic equations: 
5x2- 3a;2/  +  32  2/2  =0         ^y-x''  =  ll         Zx-2y^  =  l 

Linear-Quadratic  Pairs 

222.  Model  A.  —  A  certain  rectangular  field,  containing 
480  square  rods,  requires  104  rods  of  fence  to  inclose  it. 
What  are  its  dimensions? 


I.   Let  X 

=  the  number  of  rods  in 

length. 

mu      480 
Then  — 

=  the  number  of  rods  in 

breadth. 

® 

2.  +  2(^^«)=104 

® 

rc2  +  480  =  52  X 

®  X 

X  ■ 

^  2 

®        :» 

;2  -  52  X  +  480  =  0 

@- 

52 

X 

and  so  on. 

Or  otherwise. 

11.  Let 

X  =  the  number  of  rods 

in  length. 

Then  52 

—  X  =  the  number  of  rods 

in  breadth. 

©        X  (52  -  x)  =  480 

® 

52  a;  -  a;2  =  480 

©  same 

values 

® 

0  =  x2  -  52  a;  +  480        0  - 

52 

X+X^ 

and  so  on. 

200 


LINEAR-QUADRATIC  PAIRS  201 

Still  another  way  would  be  to  use  two  unknown  letters,  as  follows: 
III.   Let  X  =  the  number  of  rods  in  length; 
y  =  the  number  of  rods  in  breadth. 

©  xy  =  480 

@  2  X  +  2  ?/  =  104 

®  x  +  y  =  52  (2)  ^2 

Equations  ®  and  ®  we  cannot  solve  by  the  method  of 
elimination  heretofore  tried.  We  can  find  the  value  of  x 
from  the  third  equation,  —  not  the  numerical  value  that 
satisfies  both  ®  and  ®,  but  an  explicit  formula  for  x.  We 
may  then  substitute  in  the  other  equation;  that  is,  we  can 
use  in  the  first  equation  a  formula  for  x  instead  of  x  itself. 
Then,  if  the  formula  for  x  had  no  letter  x  in  it,  we  shall  have 
an  equation  from  which  x  has  been  eliminated.     Thus: 

®  X  =  52  -  y  (D  -  y 

®      y  (52  -  y)  =  480  0  subst.  in  ® 

(e)        52y  —  y^  =  480  ®  same  values 

0  0  =  2/2  -  52  2/  +  480  ®  -52  y  +  y^ 

and  so  on. 

Or  we  could  in  the  same  way  have  eliminated  y.  The  use 
of  a  second  unknown  letter  may  often,  as  in  this  case,  be  a 
way  of  explaining  how  to  get,  with  one  letter,  abbreviations 
for  two  unknown  numbers. 

The  method  just  given  is  the  one  by  which  elimination  should  usually 
be  done  in  equations  of  this  kind,  but  the  valuie  of  one  letter  may  be 
found  from  the  quadratic  equation  and  substituted  in  the  linear;  as: 

^  480  ^ 

®  X  =-. —  0-^2/ 

y 

480 
®  [•  y  =  52  0  subst.  in  ® 

®  480  +  2/2  =  52  2/  ®  X  ^ 

0        2/2  -  52  2/  +  480  =  0  ®  -  52y 

223.  Model  B.  —  A  cistern  can  be  filled  by  two  pipes, 
running  together,  in  2  hours,  55  minutes.  The  larger  pipe 
by  itself  will  fill  the  cistern  2  hours  sooner  than  the  smaller 


202  ELIMINATION  BY  SUBSTITUTION 

pipe  by  itself.     How  long  will  it  take  the  larger  pipe  alone  to 
fill  the  cistern?    The  smaller  pipe? 

Let  X  =  the  number  of  hours  for  the  smaller  pipe. 
y  =  the  number  of  hours  for  the  larger  pipe. 
(1)     x-y  =  2 


® 

x^  y      35 

® 

S5y  +  S5x  ^  12xy 

(2)  xS5xy 

® 

x^y  +  2 

®  +  y 

® 

S5y  +  S5(y  +  2)  =  12yiy  +  2) 

®  subst.  in  ® 

® 

701/ +  70  =  12y^+2^y 

0  same  values 

® 

0  =  12  7/2  -  46  2/  -  70 

®  -70  2/ -70 

® 

0  =  6  2/2  -  23  1/  -  35 

®-2 

Then  y  =  5,  or  — |. 

Arts.    5  hours  for  the  larger  pipe,  7  for  the  smaller. 
Check:    5  hours  =  7  hours  —  2  hours. 

^  of  the  tank  +  |  of  the  tank  =  ^f  of  the  tank,  in  1  hr. 
2  hrs.  55  min.  =  ff  hrs.;  ff  X  B  =  1- 

224.  Model  C.  —  A  number  of  two  digits  is  9  less  than 
the  square  of  the  sum  of  its  digits;  and  if  45  is  subtracted 
from  the  number,  the  digits  are  interchanged. 

Let  x  =  the  tens  digit; 

y  =  the  units  digit. 
Then      lOx  +  y  =  the  number. 

(a?  +  yY  =  the  square  of  the  sum  of  the  digits. 

lOy  +  X  =  the  number  with  digits  interchanged. 


®     10  X  +  2/  +  9    =  x2  +  2  xi/  +  ?/2 

@     10a;  +  i/-45  =  10  2/  +  a; 

®      9x-9i/  =  45 

®  -  IO2/-X  +  45 

®           X  -  y  =  5 

®  ^9 

®                   x=y+5 

®+l/ 

(6)     10  (1/  -h  5)  +  2/  +  9 

=  y^+10y-\-25+2y  (y+5)+y^ 

®  subst.  in  ® 

®     Uy  +  69  =  47/2  +  20^  +  25 

®  same  values 

®     0  =  42/2 +  91/ -34 

@  -  11 2/  -  59 

Then   y  ^  2,  y  =  -V". 

a;  =  2/  +  5  =  7. 

Am.    ' 

Check:  72  =  (9)^  -  9 

72  -  45  =  27 

72. 


LINEAR-QUADRATIC  PAIRS  203 

225.  The  rule  for  this  method,  which  is  called  elimination 
by  substitution,  is :  Find  an  explicit  formula  for  x  (or  for  y) 
from  the  linear  equation^  and  substitute  it  in  the  quadratic 
equation. 

Both  in  Model  B  and  in  Model  C  we  used  in  the  answer 
only  that  pair  of  values  which  was  easily  interpreted  in  the 
problem.  In  general,  however,  when  two  values  of  x  and  y 
occur,  both  pairs  should  be  given  in  the  answer,  if  there  is 
no  implied  condition  in  the  problem  making  either  of  them 
unreasonable. 

EXERCISE  77 
Solve: 

2.  xy  =  1]  Zx  —  by  =2 

3.  a;-^±l=_l;  a;2  +  i/2=25 

4.  7  a;2  -  8  x?/ =  159;  5  a;  +  ?  2/ =  7 

5.  X  -  4 2/  =  -8;  x2  +  2  2/2  =  34 

6.  3  a:2  =  336  -  4  2/2;  X  +  2/  +  14  =  0 

7.  x2  -  2  X2/  +  2  2/2  =  58;  a;  +  2/  =  13 

8.  a;2  +  5  2/2  =  4  a;2/  +  13;  3  X  -  10  2/  =  1 

9.  3  2/  -  2  a;  =  6;  4  0:2  =  9  (2/2  +  12) 

^°-   2  +  1  =  1'^  +  ^  =  ^ 


11.  3x2-  2x2/  =  15;  2  a; +  3  2/  =  12 

12.  -+^=5;  3X-22/  =  16 
2/      X      2'  ^ 

13.  x2  +  24x2/  =  (3 2/ +  4)2 +  8;  x  =32/  +  2 

'^•"^  2       -       2'^        x  +  2-^ 

15    3x- lOi/ =  1;  x2-x2/ =52/2  +  79 


204  ELIMINATION  BY  SUBSTITUTION 

16.  11 2/  +  6  a:  =  62;  x^  -  xy  +  y"^  =  13 

17.  -3-  -  -/  =  ^  ;  a;  +  2/  +  3a^2/  =  93 

18.  ^±^-^:^=Z;  x2  +  6x2/  +  2|/2H-4x+32/=29 

19.  x2  -  2  XT/  -  2/2  =  31;  x  +  2/  =  13 

20.  10-^tl^=^;  3x2-60^  +  52/  =20 

21.  2^+A_li  =  ^;  x2  +  4x^  =  32/2  +  20x  +  6 

x-y-l  y 

23.  Find  two  numbers  whose  sum  is  4  times  their  differ- 
ence, and  the  difference  of  whose  squares  is  196. 

24.  A  path  around  a  rectangular  garden  is  7  feet  wide  and 
1806  square  feet  in  area.  The  area  of  the  path  is  994  square 
feet  less  than  the  area  of  the  garden.  Find  the  area  of  the 
garden. 

25.  A  number  is  7  times  the  sum  of  its  two  digits;  and  if 
the  number  is  multiplied  by  its  first  digit,  the  product  is  672. 
Find  the  number. 

26.  The  difference  of  two  numbers  is  y\  of  the  greater 
number,  and  the  difference  of  their  squares  is  380.  Find  the 
numbers. 

27.  If  a  certain  rectangular  field  were  75  rods  longer  and 
20  rods  wider,  its  length  would  be  double  its  breadth,  and 
its  area  would  be  double  what  it  is  now.     Find  its  dimensions. 

28.  In  a  certain  block  containing  exactly  similar  houses, 
there  are  300  rooms.  There  are  5  more  houses  in  the  block 
than  there  are  rooms  in  one  of  the  houses.  How  many 
houses  are  there  in  the  block? 


LINEAR-QUADRATIC   PAIRS  205 

29.  The  front  wheel  of  a  bicycle  makes  16  turns  less  than 
the  hind  wheel  in  going  a  mile.  If  the  front  wheel  were  6 
inches  more  in  circumference,  it  would  turn  60  times  less 
than  the  hind  wheel  in  going  a  mile.  Find  the  circumference 
of  each  wheel. 

30.  When  a  certain  train  has  traveled  5  hours,  it  is  still 
60  miles  short  of  its  terminus.  One  hour  can  be  saved  on 
the  whole  trip  by  running  5  miles  an  hour  faster.  Find  the 
speed  of  the  train,  and  the  length  of  the  trip. 

31.  A  and  B  start  on  a  road  race  together.  A  is  a  sure 
winner;  looking  back  once  on  the  road,  he  sees  B  60  rods 
behind.  A  crosses  the  line  4  minutes  after  that,  and  B  comes 
in  6  minutes  behind  A.  When  A  looked  back,  he  had  as 
far  to  ride  as  B  had  already  ridden.     Find  the  speed  of  each. 

32.  The  fore  wheel  of  a  carriage  turns  132  times  more 
than  the  hind  wheel  in  going  a  mile.  6  turns  of  the  fore 
wheel  cover  2  feet  less  than  5  turns  of  the  hind  wheel.  Find 
the  circumference  of  each  wheel. 

33.  I  rowed  24  miles  downstream  and  back  again.  I  took 
8  hours  longer  on  the  return  trip,  because  the  current  reduced 
my  speed  to  J  of  what  it  was  going  downstream.  Find  the 
rate  of  the  current. 

34.  Two  trains  start  at  the  opposite  ends  of  a  double- 
track  railroad  300  miles  long.  After  they  pass,  one  train 
takes  9  hours,  the  other  4  hours,  to  complete  the  journey. 
Find  the  speed  of  each  train. 

35.  If  1  is  added  to  the  denominator  of  a  certain  fraction, 
the  value  of  the  fraction  becomes  J.  If  2.1  is  added  to  the 
fraction,  the  fraction  is  inverted.     Find  the  fraction. 

36.  If  2  is  added  to  the  denominator  of  a  certain  fraction, 
the  value  of  the  fraction  becomes  ^.  If  4  is  added  to  each 
term  of  the  fraction,  its  value  is  increased  by  .125.  Find 
the  fraction. 


206  ELIMINATION  BY  SUBSTITUTION 

37.  A  rectangular  field  238  square  rods  in  area  loses  58 
square  rods  because  a  strip  one  rod  wide  is  taken  off  all 
around  the  edge  for  roads.  Find  the  dimensions  of  the  field 
that  remains. 

38.  Five  leaks  and  a  drainpipe  empty  a  cistern  in  4  hours. 
The  average  time  for  one  of  the  leaks  to  empty  the  cistern 
would  be  24  hours  more  than  half  the  time  the  drainpipe 
requires.  How  long  will  it  take  to  empty  the  cistern  if  the 
leaks  are  stopped? 

39.  When  a  certain  kind  of  cloth  is  wetted,  it  shrinks  J 
in  length  and  tV  in  width.  A  piece  of  this  cloth  used  for 
a  rectangular  awning  was  found  to  have  shrunk  8f  square 
yards  in  area.  Then  the  edging  bought  for  it  before  it  shrank 
(intended  to  go  on  one  long  side  and  both  ends)  was  found  to 
be  2  yards  too  long.  What  were  the  length  and  the  width 
of  the  cloth  before  shrinking? 

40.  The  area  of  a  rectangular  plot  of  land  is  16  square 
feet  less  than  a  square  plot  of  equal  perimeter;  and  its 
breadth  is  1  foot  more  than  |  of  its  length.  Find  the  di- 
mensions of  the  rectangular  plot. 

41.  Two  boys  run  in  opposite  directions  around  a  rec- 
tangular field,  the  area  of  which  is  one  acre.  They  start  at 
one  corner  and  meet  13  yards  from  the  opposite  corner; 
and  one  of  the  boys  could  go  around  6  times  while  the  other 
was  going  5  times.     Find  the  dimensions  of  the  field. 

An  acre  =  160  square  rods.  A  square  rod  is  a  square  5.5  yards  on 
a  side. 

Algebraic  Diagrams 

226.  The  algebraic  diagram  for  a  two-letter  quadratic 
equation  cannot  be  a  straight  line;  for,  as  we  learned  in 
Chapter  IX,  a  straight  line  is  always  the  locus  of  a  two- 
letter  equation  of  the  first  degree.  To  get  a  diagram  for  a 
two-letter  quadratic,  we  must  use  the  equation  as  an  im- 


ALGEBRAIC  DIAGRAMS  207 

plicit  formula  in  which  we  can  give  a  series  of  values  to  one 
letter,  in  order  to  find  the  corresponding  value  (or  values) 
of  the  other. 

227.  The  following  example  furnishes  one  of  the  simplest 
illustrations  of  this  method  of  plotting  a  two-letter  quad- 
ratic equation: 

Model  D.  —  The  diagonal  of  a  rectangle  is  5  inches  long. 
The  perimeter  of  another  rectangle  of  the  same  width  but  twice 
as  long  is  22  inches.    Find  the  dimensions  of  the  first  rectangle. 

Let  X  =  number  of  inches  in  the  length  of  the  rectangle; 

y  =  number  of  inches  in  the  width  of  the  rectangle. 
Then      x^  -{-  y^  =  the  square  of  the  diagonal. 

4x  +  2y  =  the  perimeter  of  the  other  rectangle. 
0  x2  +  ?/2  =  25 

@      4:x  +  2y  =  22 

®         2x-\-y  =  11  0^2 

0  y  =  11  -2x  0  -  2a; 

©  a;2  +  (121  -  44  re  +  4  x2)  =  25         0  subst.  in  0 

0  5a;2-44x  +  96  =  0  0-25 

,    ©  (5  X  -  24)  (x  -  4)  =  0  0  factored 

Equation  ©  leads  to  two  solutions,  one  giving  a  rectangle  4  inches 
by  3  inches,  the  other  a  rectangle  4.8  inches  by  1.4  inches. 

In  drawing  the  two  loci  that  form  the  algebraic  diagram 
for  this  example,  0  gives  us  a  straight  line  with  intercepts 
5|  and  11.  We  must  consider  ®  an  implicit  formula,  so 
that  we  can  give  a  series  of  values  to  one  letter  and  find  the 
corresponding  values  of  the  other. 

Thus  we  find  that  when  x  =  1,  1  -{-  y^  =  25,  and  y^  =  24. 

Whence,  y  =  4.9  and  y  =  —4.9. 

We  shall  have  the  same  values  if  a;  =  —1. 

In  that  way  we  can  construct  a  table  as  follows: 
X  1  1  2  233445 

y       4.9       -4.9      4.6       -4.6      4-4      3-3      0 
When  X  is  greater  than  5,  y^  becomes  negative,  and  con- 
sequently we  cannot  find  a  corresponding  value  of  y. 


208 


ELIMINATION  BY  SUBSTITUTION 


228.  If  we  now  plot  the  values  in  the  above  table,  we 
shall  find  that  they  lie  on  a  semicircle,  to  the  right  of  the 
2/-axis,  with  its  center  at  the  point  where  the  axes  cross 
{the  origin).     If  we  plot  the  values  of  y  corresponding  to  the 

negative  values  of  x,  we  shall 
find  them  lying  on  the  other 
half  of  the  same  circle.  The 
locus  of  equation  ©,  then,  is 
a  circle  with  radius  5,  having 
its  center  at  the  origin. 

229.  With  our  knowledge  of 
geometry  we  can  prove  that 
the  circle  just  described  is  the 
^^^-  ^'  required  locus  of  equation  ®; 

for  the  equation  x^  +  y^  =  25  simply  means  that  the  slant 

side  OP  of  the  right  triangle  OMP  (where  P  is  the  point 

represented  by  x  and  y) 

must  always  be  equal  to  5. 

Consequently  the  point 

P  will  have  a  constant 

distance  5  from  the  origin. 

The  Equation  of  the 
Circle 

230.  Find  the  equation 
of  a  circle  with  radius  10, 
having  its  center  at  the 
point  which  represents 
the  number-pair  5,7. 

The  "number-pair  5,7" 
means  the  pair  of  values 
a;  =  5  and  y  =  7. 

Every  point  on  the  circle  will  represent  a  pair  of  numbers. 
Thus  P  represents  the  numbers  x  and  y,  of  which  a;  is  de- 


FiG.  2. 


ALGEBRAIC  DIAGRAMS  209 

fined  by  the  length  of  OK,  and  y  by  the  length  of  KP,  just 
as  the  length  of  OL  is  5  units  and  the  length  of  LC  is  7  units. 
Since  P  is  on  the  circle,  CP  must  be  equal  to  10  units;  and 
that  gives  us  the  equation 

{x  -  5)2  +{y-  7)2  =  100     (Why?) 

This  will  be  true  whenever  P  is  on  the  circle.  Even  when 
X  is  less  than  5,  making  x  —  b  negative,  the  square  {x  —  5)^ 
will  be  positive.  In  the  same  way,  {y  —  7)^  will  always  be 
positive.  We  shall  be  concerned,  then,  only  with  the  differ- 
ence of  y  and  7,  not  with  its  sign.  For  every  point  on  the 
circle,  therefore, 

©     {x  -  5)2  +  {y-  7)2  =  100 

®    a;2  -  10  X  +  25  +  2/2  _  14  y  _}_  49  =  100       Q  same  values 

®    x2  +  2/2  -  10  X  -  14  ?/  -  26  =  0  0-100 


EXERCISE  78 

1.  Find  the  equation  of  the  circle  with  center  at  3,2  and 
the  radius  5. 

2.  Find  the  equation  of  the  circle  with  center  at  5,0  and 
the  radius  3. 

3.  Find  the  equation  of  the  circle  with  the  center  at  0,3 
and  the  radius  8. 


Find  the  equations  of  the  circles  having; 

4.  Center  at  3,5  and  radius  6. 

5.  Center  at  —2,4  and  radius  7. 

6.  Center  at  5,-3  and  radius  4. 

7.  Center  at  —4,-3  and  radius  9. 

8.  Center  at  7,— 2  and  radius  11. 

9.  Center  at  —5,8  and  radius  3. 
10.  Center  at  4,— 4  and  radius  4. 


^' 


c-^-. 


210  ELIMINATION  BY  SUBSTITUTION 

11.  Find  the  equation  of  the  circle  with  center  at  the 
origin,  and  radius  17.  Where  is  this  circle  intersected  by 
the  straight  line  that  passes  through  the  points  3,20  and 
11,12? 

12.  Find  the  equation  of  the  circle  that  has  its  center  at 
the  point  2,5  and  its  radius  equal  to  5.  Where  is  this  circle 
intersected  by  the  line  that  cuts  each  axis  14  units  from  the 
origin? 

Other  Quadratic  Loci 

231.  The  equations  of  circles  that  we  have  thus  far 
found  have,  for  terms  of  the  second  degree,  the  sum  of  the 
squares  of  x  and  y.  We  have  not  yet  proved,  though  it  is 
true,  that  every  equation  of  this  type  represents  a  circle. 
Quadratic  two-letter  equations  of  other  types,  however, 
are  very  common.  Let  us  consider  first  those  appearing  in 
the  model  problems  of  this  chapter. 

In  Model  A,  we  have  the  equation 

xy  =  480 

and  we  take  this  as  an  implicit  formula  for  y,  using,  only  such 
values  of  x  as  make  computation  easy  and  give  points  con- 
veniently placed  in  the  diagram: 


x 

8  10  12  15  16  20  24  30  32  40  48 

60 

y 

60  48  40  32  30  24  20  16  15  12  10 

8 

If  we  changed  the  sign  of  each  value  of  x,  we  should  have 
the  negatives  of  the  same  values  of  y.  The  points  thus 
represented  would  lie  in  the  lower  left-hand  quarter  of  the 
diagram. 

Drawing  the  line  represented  by  the  linear  equation  of 
Model  A,  we  find  that  the  loci  intersect  in  two  points  which 
correspond  to  the  two  solutions  of  the  equations. 


.HiA 


.-^  ..Ac^r^ir 


Jo^^^ 


ALGEBRAIC  DIAGRAMS 


211 


"^~ 

■^~" 

> 

V 

-80 

"^■~ 

\ 

s 

\ 

\ 

I 

V 

\ 

s> 

in 

s 

^ 

k^ 

>!o 

^ 

s50 

■ ' 

-^ 



\ 

V 

^ 

s 

\ 

V 

\ 

V 

\ 

\ 

\ 

Fig.  3. 


232.  In  Model  B,  equation  0  gives  for  each  value  of  x 
one  value  of  y,  just  as  in  Model  A;  but  in  this  case  the 
negative  values  of  x  give  totally  different  values  of  y.  Thus 
we  have,  for  the  value  x  =  10,  the  following  equation  for  y: 

35  2/  + 350  =  1201/ 

from  which  we  obtain. i/  =  4.0. 

On  the  other  hand,  for  the  value  x  =  — 10,  the  equation 
becomes 


®    35  2/- 350  =  -1202/ 
@  155  2/  =  350 

from  which  we  obtain  y  =  2.3. 


©  +  1202/ +  350 


212 


ELIMINATION  BY  SUBSTITUTION 


The  table  of  corresponding  values  is: 

a:      -10     -7    -5    -2    -1     0  1  2        3  4      5 

y        2.3     2.0     1.8     1.3       .7     0     -1.5     -6.4     105     10.8     7 

We  can  plot  the  locus  of  x  —  y  =  2  and  find  the  two 
points  of  intersection  of  the  loci. 


7  10 
5    4 


Fig.  4. 

233.  In  Model  C,  the  equation  we  have  for  an  implicit 

formula  is 

10  X  +  2/  +  9  =  x2  +  2  a;?/  +  2/2 

and  when  we  substitute  successive  values  of  x  we  get  in 
each  case  a  quadratic  for  y,  which  would  generally  have  two 
roots.    Thus  we  get  the  following  table: 

X  Quadratic  for  y  Values  of  y 

0  2/2-2/-9=0  3.5,  -2.5 

1  2/2  +  y  -  18  =  0  3.8,  -4.8 

2  2/2 +  32/- 25  =0  3.7,  -6.7 

3  i/2  +  5i/-30  =  0  3.5,  -8.5 
5  2/2 +  9?/- 34  =0  2.9,  -11.9 
7  2/^  +  13  2/  -30  =0  2,-15 

-1  2/' -3  2/ +  2=0  1,2 

-2  2/^  -  5  2/  +  15  =  0  none 

-3  2/^^  -  7  2/  +  30  =  0  none 

-.5  2/2  -  2  2/  -  3.75  =  0  3.2,  -1.2 


ALGEBRAIC  DIAGRAMS 


213 


By  plotting  the  values,  we  get  the  curve  given  in  the  fol- 
lowing diagram.  This  curve  is  cut  by  the  line  x  —  y  =  5  in 
the  point  7,2,  and  also  in  the  point  |,— 4j. 


Fig.  5. 


^ 

^ 

■^ 

/^ 

-— ^ 

s 

I 

^ 

1 

^ 

Ji^ 

^ 

S 

£ 

;         ' 

:           . 

)      I 

\ 

\ 

/ 

^ 

^ 

— 

Fig.  6. 

234.  The  equations  of  Ex.  5,  Exercise  77,  give  the  diagram 
of  Fig.  6. 


214  ELIMINATION  BY  SUBSTITUTION 

235.  The  curves  obtained  for  Models  A  and  B  are  called 
h3rperbolas.  Each  of  these  curves  consists  of  two  branches, 
and  does  not  inclose  a  space. 

The  curve  obtained  for  Model  C  is  a  curve  of  one  branch ; 
this  also  is  open,  that  is,  it  does  not  inclose  a  space.  It  is 
called  a  parabola. 

The  curve  shown  in  Fig.  6  is  a  closed  curve,  not  all  the 
diameters  of  which  are  equal,  as  in  a  circle;  it  could  be 
described  as  a  circle  shrunk  crosswise.  It  is  called  an 
ellipse. 

A  quadratic  locus  can  never  be  a  single  straight  line,  but 
it  can  be  a  pair  of  straight  lines. 

Such  figures  as  these  form  the  subject  matter  of  a  very 
interesting  study  called  Analytic  Geometry.  For  our  pur- 
pose here  it  is  enough  to  notice  that  the  locus  of  a  quad- 
ratic two-letter  equation  can  generally  be  cut  by  a  straight 
line  in  two  points. 

EXERCISE  79 
Plot  ihe  following  loci:  r 

1.  (Circle)  x"  +  y^  =Sx-\-Qy 

2.  (Circle)  x"  +  y^  -  S  x  =  2S 

3.  (Pair  of  straight  lines)  24  y^  —  25  x^  =  25  xy 

4.  (Parabola)  2/'  =  4  x  -  20 

5.  (Parabola)  S  {x  -  2yy  =  7 x  +  Q2 

6.  (Parabola)  x^  -  Qxy  +  9y^  =  Ax  +  4S 

7.  (EUipse)  x2  +  3  2/2  =  5  X  +  18 

8.  (EUipse)  x^  +  xy  +  y^=  28 

9.  (Hyperbola)  2x^  —  xy  —  Qy^  =28 
10.   (Hyperbola)  5  x^  —  3  xt/  =65 


ALGEBRAIC  DIAGRAMS  215 

The  standard  Parabola 

236.  The  locus  of  the  equation  y  =  x^  is  called  the 
Standard  Parabola.     It  is  plotted  as  follows  (Fig.  7) : 

\     X     :    0    ±1     ±2     ±3     ±4     ±5     ±6     ±7     ±8 
\      y         0        1        4        9      16      25      36      49      64 

237.  The  standard  parabola  and  its  equation  can  be 
utilized,  as  will  now  be  shown,  for  a  graphical  check  on  the 
solution  of  a  one-letter  quadratic  equation. 

Consider  the  equation  x^  -\-  5  x  =  7. 

This  can  be  written  x^  =  —  5  x  +  7. 

This  equation  will  be  satisfied  by  any  value  of  x  that  will 
make  x^  and  —  5  a;  +  7  equal;  that  is,  by  any  value  of  x 
that  will  give  the  same  value  of  y  in  the  two  explicit  formulas 
y  =  x^  and  y  =  —  5  a;  +  7. 

The  first  of  these  formulas  is  the  standard  parabola,  and 
the  other  is  a  straight  line  having  an  intercept  1.2  on  OX 
and  7  on  OY.  Where  these  two  loci  intersect,  the  same 
number-pair  will  satisfy  both  formulas. 

Our  check,  then,  is  as  follows:  On  a  carefully  drawn  dia- 
gram of  the  standard  parabola,  lay  a  straight  line  (a  ruler- 
edge,  or  a  line  ruled  on  tracing  paper)  through  the  intercepts 
of  the  line  y  =  —  5  x  +  7.  Either  point  where  this  line 
crosses  the  parabola  will  represent  a  value  of  x  which  gives 
the  same  value  of  y  in  both  loci.  This  value  of  x  is,  there- 
fore, a  root  of  the  equation  x^  +  5  x  =7. 

238.  In  the  same  way,  the  equation  5  x^  —  14x  +  2  =0 
can  be  written  x^  =  2.8  x  —  .4.  Its  roots  will  be  indicated 
by  the  intersections  of  the  standard  parabola  with  the 
straight  line  y  =  2.8  x  —  A,  or  5y  =  14  x  —  2. 

It  is  easy  to  see  that  the  straight  line  will  sometimes 
touch  the  standard  parabola  in  only  one  point.    What  may 


216  ELIMINATION  BY  SUBSTITUTION 


50 

\ 

OK 

1              "n 

u 

/ 

\ 

4u\ 

20            -1 

5             -1 

0 

S 

0'              ■ 

)          1 

0              1 

5               2( 

Fig.  7. 


ALGEBRAIC  DIAGRAMS  217 

we  then  conclude  about  the  roots  of  the  equation?  Also, 
the  line  might  miss  the  parabola  altogether.  Then  we  know 
that  the  roots  of  the  equation  are  imaginary. 

EXERCISE  80 

By  using  the  diagram  of  the  standard  parabola,  Fig.  7, 
find  roughly  the  roots  of  the  following  equations,  and  verify 
them  by  the  algebraic  solutions. 

1.  a;2=x  +  30  6.  2a;2-16=3x 

2.  x2  -  15  =  2  X  7.  7  x2  =  32 

3.  2  x2  +  X  =  15  8.  2  x2  -  9  a;  +  14  =  0 

4.  x2  =  16  a;  -  20  9.  2  x^  -  19  x  +  14  =  0 

5.  4a;2-28a;  =49  lo.  lla;^  +  13a;  +  3  =  0 


CHAPTER   XII 

SUPPLEMENTARY     PROBLEMS     FOR     PRACTICE 
AND  REVIEW 


Invent   algebraic   expressions   for   the   following   sets   of 
numbers. 

1.  Any  two  numbers;  any  three  numbers. 

2.  The  sum  of  any  two  numbers;    the  difference  of  any 
I  wo  numbers. 

3.  The  product  of  any  two  numbers;  the  product  of  any 
three  numbers. 

4.  Any  two  consecutive  numbers;  any  three. 

5.  The  product  of  any  two  consecutive  numbers. 

6.  The  product  of  any  three  consecutive  numbers. 
7".   The  sum  of  any  two  consecutive  numbers. 

8.  The  sum  of  any  three  consecutive  numbers. 

9.  Any  odd  number;  any  even  number. 

10.  Any  two  odd  numbers;  any  two  even  numbers. 

11.  Any  two  consecutive  odd  numbers;  their  sum;  their 
product. 

12.  Any  two  consecutive  even  numbers;  their  sum;  their 
product. 

13.  Any  three  consecutive  odd  numbers;  their  sum;  their 
product. 

14.  Any  three  consecutive  even  numbers;  their  sum;  their 
product. 

218 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         219 

15.  The  square  of  any  number;  the  squares  of  any  two 
numbers. 

16.  The  sum  of  the  squares  of  any  two  numbers. 

17.  The  difference  of  the  squares  of  any  two  numbers. 

18.  The  square  of  the  sum  of  any  two  numbers. 

19.  The  square  of  the  difference  of  any  two  numbers. 

20.  The  difference  of  the  squares  of  two  consecutive  odd 
numbers. 

21.  The  difference  of  the  squares  of  the  two  different 
numbers  formed  by  the  same  two  digits;  the  square  of  their 
difference. 

II 

1.  What  is  the  supplement  of  X  degrees? 

2.  What  is  the  complement  of  3  X  degrees? 

3.  What  is  the  complement  of  Y  degrees?        : 

4.  What  is  the  supplement  of  5  F  degrees? 

6.  Two  angles  have  a  ratio  3.  If  the  number  of  degrees 
in  the  smaller  is  represented  by  A,  how  many  degrees  are 
there  in  their  sum?     In  the  complement  of  their  sum? 

6.  Two  angles  have  a  ratio  11.  If  the  number  of  degrees 
in  the  smaller  is  represented  by  B,  how  many  degrees  are 
there  in  the  supplement  of  their  sum? 

7.  Two  angles  have  a  ratio  4.  Represent  the  number  of 
degrees  in  the  smaller  by  some  letter,  and  obtain  a  way  of 
expressing,  first,  their  difference;  and  then,  the  comple- 
ment of  their  difference. 

8.  What  is  the  complement  of  C  degrees?  Of  C  +  1 
degrees?  Of  C  +  2  degrees?  What  would  these  three 
values  be  if  the  value  of  C  were  20? 

9.  What  is  the  supplement  of  D  degrees?  Of  D  —  1 
degrees?  Of  D  —  2  degrees?  What  would  these  three 
values  be  if  the  value  of  D  were  68? 


220 


PROBLEMS  FOR  PRACTICE  AND  REVIEW 


10.  What  is  the  complement  of  A  +  20  degrees?  Of 
A  —  33  degrees? 

11.  What  is  the  supplement  of  X  +  30  degrees?  The 
complement  of  X  —  25  degrees? 

12.  What  is  the  complement  of  2  X  —  3  degrees?  The 
complement  of  3  X  —  2  degrees?  The  supplements  of  the 
same  angles? 

13.  If  two  angles  have  a  ratio  5,  how  would  you  repre- 
sent them?  If  the  angles  were  supplements,  how  could  you 
state  that  fact  by  means  of  an  equation? 

14.  In  the  stripe  in  Fig.  1,  the  angle  A  is  8°  less  than  X; 
A  and  Y  have  the  ratio  3.    How  many  degrees  are  there  in  X? 


Fig.  1.  Fig.  2. 

15.  In  the  stripe  in  Fig.  2,  the  angle  X  is  12°  less  than 
three  times  the  angle  7.    How  many  degrees  are  there  in  X? 

16.  In  Fig.  3,  A  =  38.3°,  C  =  53.6°.    How  large  is  B? 


Fig.  4. 


Fig.  3. 

17.  In  Fig.  4,  Z  =  56.4°,  X  and  Y  have  the  ratio  2. 
How  large  is  X? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW        221 


18.  In  Fig.  5,  the  ratio  of  Q  to  P  is  1.6,  and  R  is  24°  less 
than  their  sum.    How  large  is  each  angle? 

19.  In  Fig.  6,  A  =  C  —  5,  and  the  ratio  of  A  to  5  is  ^. 
How  large  is  each  angle? 


Fig.  5.  Fig.  6.  Fig.  7. 

20.  In  Fig.  7,  X  is  35°  less  than  7,  and  Z  is  2°  greater 
than  Y.    How  large  is  each  angle? 

21.  In  each  of  the  following  stripe  diagrams,  how  many 
degrees  does  A  represent? 

^    ''      U+30 


Fig.  8. 
22.  In  each  of  the  following  diagrams,  how  many  degrees  are 
there  in  the  angle  X? 


Fig.  9. 


222    PROBLEMS  FOR  PRACTICE  AND  REVIEW 

23.  The  ratio  of  the  angles  at  the  base  of  a  triangle  is  If; 
and  the  ratio  of  the  angle  at  the  vertex  to  the  larger  base 
angle  is  1.4.     Find  the  angles  of  the  triangle. 

Let  one  angle  equal  180°  minus  the  sum  of  the  other  two  angles. 

24.  The  ratio  of  the  two  smaller  angles  of  a  triangle  is 
1.33;  and  the  ratio  of  the  largest  angle  to  the  smallest  is 
2.17.     Find  the  angles  of  the  triangle. 

Ill 

Convert  into  Fahrenheit  the  following  temperatures 
Centigrade : 

1.  -80°  C.  6.  20°  C. 

2.  -40°  C.  6.     31.32°  C. 

3.  0°C.  7.     98.16°C.      . 

4.  4°C.  8.     458. 1°C. 

9.   33.44° C. 

Convert  into  Centigrade  the  following  temperatures 
Fahrenheit : 

10.  0°F.  14.      -40°  F. 

11.  16°  F.  15.      -80°  F. 

12.  -15°F.  16.         68°  F. 

13.  32°  F.  17.   91.56°  F. 

18.  What  temperature  Fahrenheit  is  just  double  the  cor- 
responding temperature  Centigrade?  What  temperature 
Centigrade  is  just  double  the  corresponding  temperature 
Fahrenheit? 

19.  What  temperature  has  the  same  number  of  degrees 
in  both  Centigrade  and  Fahrenheit  scales? 

20.  A  certain  temperature,  observed  with  Centigrade  and 
Fahrenheit  thermometers,  is  recorded  by  writing  numbers 
that  differ  by  60.    What  are  the  two  thermometer  readings? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         223 

IV 

Solve: 

1.  7  (19  X  -  21)  =  11  (12  X  -  3)  -  30 

2.  9  (25  x  +  11)  =  19  (12  a;  -  5)  +  2 

3.  51  X  -  1135  -f  13  (103  -x)  =  11  (2  X  +  20) 

4.  203  -  17  X  =  3  (x  -  505)  -  2 

6.  1170  +  (x  -  4)  =  11  (x  -  3)  +  22 

6.  9  (103  X  -  820)  =  206  (2  -  x)  +  139 

7.  800  {x  -  10)  +  2  (11  X  -  328)  -  1  -  35  x  =  0 

8.  3000  (x  -  5)  =  10  (100  -  27  x)  H-  355  -x 

9.  10  (2  X  -  3)  =  2  (x  +  15)  +  6 

10.  3(3x-  ll)  +  2(5x  +  2)  =5  +  9(2x-5) 

11.  5  (7  X  -  2)  +  2  (5  X  -  13)  =  3  (11  x  +  2)  +  26 

12.  17  (3x-  l)+3(5x-  17)  =  ll(5x-3)  +  x  +  5 

13.  13  (x  -  11)  +  17  (2  X  -  27)  -  1  =  3  (x  -  29) 

14.  101  (5  X  +  17)  =  203  (3  X  +  23)  +  3  X  -  3059 

15.  4  (x  -  13)  =  13  (x  -  4)  -  81 

16.  117  (x  -  25)  +  25  (x  +  117)  =  3x  +  50  +  3  (7  -  x) 

17.  8  (x  -  1)  +  4  (9  -  4x)  =  4  +  17  (3  -  x) 

18.  17  (2x  +  4)  +  13  (2x  -  5)  =  3  (x  +  17) 

19.  51  X  -  1137  +  13  (103  -  x)  =  11  (2  X  +  20)  -  2 

20.  I  (18x  -  252)  =  §  (15  -  3x)  +  1 

21.  i  (24x  -  800)  =  f  (45  -  5  x)  -  5 

22.  I  (60x  -  875)  =  i  (35  +  lOx)  -  f  X 

23.  f  (1185  X  -  5925)  =  A  (11  X  -  55)      • 

24.  i  (93x  -  138)  =  f  (lOx  -  25) 

25.  t  (56  X  +  42)  =  5  (2  X  +  16) 

26.  f  (42  X  -  48)  -  X  =  3  (x  -f  17) 

27.  I  (60  -  22  x)  +  T%  (30  -  50  x)  =0 

28.  i(5x  +  25)  +  f  (4X-28)  =f  (3x  +  9) 


224         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

29.  f  (21  a:  -  57)  =  2  (6  X  +  17) 

30.  ISx  +  3  +  t  (18  -  12a:)  =  3  (7  -  a;)  +  11  -  a; 

31.  31  (x  -  22)  +  f  (21  x  +  18)  =  2a:  +  883 

^^'        4       -      10      +20  +  ^1    5    ; 
5  a;  +  1  _  21  -  3  a;      »  /2  a;  +  3\ 


6- 


1.  A  is  12  years  older  than  B;    in  2  years  he  will  be  4 
times  as  old  as  B.     What  are  their  ages  now? 

2.  A  is  6  times  as  old  as  B;   in  4  years  he  will  be  only  4 
times  as  old  as  B.    What  are  their  ages  now? 

3.  A  is  18  years  older  than  B;    in  5  years  he  will  be  3 
times  as  old  as  B.     Find  their  ages  now. 

4.  A  is  9  times  as  old  as  B;   in  6  years  he  will  be  only  6 
times  as  old  as  B.    Ages  now? 

5.  A  is  7  years  younger  than  B;  in  4  years  he  will  be  half 
as  old  as  B.     Ages  now? 

6.  In  3  years  A  will  be  i  as  old  as  B;  now  he  is  20  years 
younger  than  B.     Ages  now? 

7.  A's  age  at  present  is  J  of  B's;   10  years  ago  it  was  }  of 
B's  age.     Ages  now? 

8.  A  was  8  times  as  old  as  B  one  year  ago;   he  is  now 
only  5  times  as  old  as  B.     Ages  now? 

0.   A  is  30  years  older  than  B;  5  years  hence  his  age  will 
be  6  times  B's.    Ages  now? 

10.  A  is  7  times  as  old  as  B;  in  10  years  he  will  be  only 
twice  as  old  as  B.    Ages  now? 

11.  The  difference  between  a  father's  age  and  his  son's  is 
28  years.     How  old  was  the  father  when  the  son  was  i  of  his 


PROBLEMS  FOR  PRACTICE  AND  REVIEW        225 

father's  age?    How  old  was  the  son  when  the  father  was  9 
times  as  old  as  the  son? 

12.  I  paid  $68  with  $10  bills  and  $2  bills,  using  10  more 
twos  than  tens.     How  many  bills  of  each  kind  did  I  use? 

13.  I  paid  $2.35  with  quarters  and  nickels,  using  5  more 
nickels  than  quarters.  Find  the  number  of  coins  of  each 
denomination. 

14.  I  paid  $139  with  fives  and  twos,  using  3  more  twos 
than  fives.     Find  the  number  of  bills  of  each  denomination. 

15.  I  paid  $7.55  with  dimes  and  quarters,  using  12  more 
quarters  than  dimes.  Find  the  number  of  coins  of  each 
denomination. 

16.  I  paid  $6.40  with  nickels  and  half-dollars,  using  4 
more  halves  than  nickels.  Find  the  number  of  coins  of  each 
denomination. 

17.  Three  times  as  many  dimes  as  nickels,  5  times  as 
many  quarters  as  dimes;  in  all  $28.70.     How  many  of  each? 

18.  A  man  borrowed  some  coins  and  paid  back  33  more 
coins  of  another  sort.  The  coins  borrowed  were  half-dollars, 
those  repaid  were  nickels;  he  still  owed  $1.50.  Find  the 
number  of  coins  borrowed. 

19.  I  sent  for  deposit  two  rolls  of  bills;  a  roll  of  twos  and 
a  roll  of  fives.  The  latter  roll  contained  50  bills  fewer,  but 
was  worth  $26  more,  than  the  first  roll.  How  many  bills 
were  there  in  each  roll? 

20.  In  a  packet  of  $10  bills  there  are  3  fewer  bills  than  in 
a  packet  of  $2  bills  which  lies  near  by.  The  first  packet  is 
worth  $74  more  than  the  second.  How  many  bills  are  there 
in  each  packet? 

21.  A  shopkeeper,  counting  his  cash,  finds  26  more  nickels 
than  quarters.  The  nickels  amount  to  10  cents  less  than  the 
quarters.     How  many  coins  of  each  kind  has  he? 


226    PROBLEMS  FOR  PRACTICE  AND  REVIEW 

22.  A  man  paid  $2.20  in  quarters  and  nickels,  using  two 
more  nickels  than  quarters.  Find  the  number  of  each  kind 
of  coin. 

23.  A  carpenter  worked  on  a  job,  and  at  the  end  of  the 
day  sent  in  his  time  as  10  hours,  although  he  had  worked 
but  8  hours.  He  had  worked  overtime,  and  for  every  hour 
overtime  he  counted  IJ  hours.  How  many  hours  did  he 
work  overtime? 

24.  A  carpenter  worked  from  2 :  30  to  8 :  30  p.m.  He  sent 
in  his  time  as  10  hours,  counting  every  hour  overtime  as  2 
hours.     How  many  hours  did  he  work  overtime? 

25.  A  plumber  worked  from  11:45  a.m.  to  8: 15  p.m.  and 
sent  in  his  time  as  10  hours,  counting  every  hour  overtime 
as  IJ  hours.     How  many  hours  did  he  work  overtime? 

26.  A  gas  fitter  worked  from  11:30  a.m.  to  9:30  p.m.  and 
sent  in  his  time  as  12J  hours,  counting  every  hour  overtime 
as  IJ  hours.     How  many  hours  did  he  work  overtime? 

27.  A  steam  fitter  worked  7|  hours  and  sent  in  his  time 
as  11  hours,  charging  overtime  double.  How  many  hours 
did  he  work  overtime? 

28.  A  plumber  worked  9  hours  and  sent  in  his  time  as  10§ 
hours,  charging  overtime  as  time  and  a  half.  How  many 
hours  did  he  work  overtime? 

29.  A  gas  fitter  spent  on  a  job  12|  hours,  and  sent  in  his 
time  as  15  hours,  charging  overtime  as  time  and  a  third. 
How  many  hours  did  he  work  overtime? 

30.  The  common  stock  of  a  gas  company  pays  4  per  cent 
and  the  preferred  stock  7  per  cent.  45  shares,  part  of  which 
is  preferred  stock,  pay  $240  per  year.  How  many  of  these 
shares  are  common  stock?    How  many  are  preferred  stock? 

31.  An  insurance  agent  proposes  to  insure  a  man  for 
$9000  by  two  limited-payment  policies,  one  at  $40  per 
thousand,   the   other  at   $35.60.      The    aggregate    annual 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         227 

premium  on  the  two  policies  is  $338.     How  many  thousand 
dollars  are  there  in  each  poUcy? 

32.  An  insurance  agent  proposes  to  insure  a  man  for 
$11,000  by  two  policies,  one  at  $38.53  per  thousand,  the 
other  at  $43.15  per  thousand,  the  aggregate  premium  be- 
ing $442.31.  How  many  thousand  dollars  are  there  in  each 
policy? 

33.  A  college  student  borrowed  $1000  and  20  years  after- 
ward he  repaid  it  with  $1080  interest.  The  interest  ran  at 
6  per  cent  for  some  years  and  was  afterward  reduced  to  5 
per  cent.  How  long  was  the  higher  rate  of  interest  in 
force? 

34.  An  investor  bought  24  shares  of  stock,  from  which 
he  received  annually  $100  in  dividends.  Some  of  the  stock 
paid  5 J  per  cent  dividends;  four  times  as  many  shares  paid 
4|  per  cent,  and  the  rest  paid  3  per  cent  dividends.  How 
many  shares  of  each  kind  did  the  investor  buy? 

35.  A  town  issued  40  $1000  bonds,  some  bearing  interest  at 
3|  per  cent,  and  the  rest  at  5  per  cent.  The  town  treasurer 
had  to  make  annual  payments  of  $1730  interest  on  these 
bonds.     How  many  bonds  of  each  kind  were  issued? 

36.  A  man  took  out  a  life-insurance  policy  when  he  was 
21  years  old,  paying  $19.62  per  thousand.  When  he  was 
30  years  old,  he  took  out  more,  paying  $24.38  per  thousand. 
At  40,  he  took  out  enough  to  double  his  insurance,  paying 
for  the  additional  pohcy  $33.01  per  thousand.  He  was  then 
carrying  $16,000  insurance,  and  paying  in  premiums  $449.60 
per  year.     What  was  the  amount  of  his  first  policy? 

37.  The  ratio  of  births  to  deaths  in  a  certain  country  is 
annually  about  1.3;  the  ratio  of  immigration  to  emigration 
is  about  17;  and  the  ratio  of  deaths  to  emigration  is  20. 
The  net  annual  increase  of  population  is  132,000.  Find  the 
annual  immigration. 


228         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

38.  I  drove  westward  in  a  carriage  for  ten  hours,  and  then 
rode  in  a  train  5  times  as  fast  for  two  days  and  nights.  The 
total  journey  was  600  miles.     How  fast  did  the  carriage  go? 

39.  Two  travelers  started  toward  each  other  from  op- 
posite ends  of  a  straight  road  132  miles  long;  one  went  4 
miles  an  hour,  the  other  7  miles.  How  long  was  it  before 
they  met? 

40.  Two  trains  start  in  opposite  directions  from  the  same 
station  at  the  same  time,  and  in  5  hours  are  337|  miles 
apart.  If  the  ratio  of  their  speeds  is  2,  what  is  the  speed  of 
each? 

41.  A  man  starts  on  a  journey  of  163  miles.  He  walks 
13  hours,  then  rides  twice  as  fast  for  20  hours,  and  finally  has 
to  stop  4  miles  short  of  his  journey's  end.  How  fast  did  he 
walk? 

42.  Two  trains  start  at  the  same  moment  from  opposite 
ends  of  a  33-mile  track,  one  going. 20  miles  an  hour,  the 
other  24  miles  an  hour.  They  stop  when  two  miles  apart. 
How  long  were  the  trains  running? 

43.  A  and  B  start  at  8  a.m.  from  opposite  ends  of  a  10- 
mile  road  and  walk  toward  each  other,  A  walking  one  mile 
an  hour  faster  than  B.  They  meet  at  9:15  a.m.  How  fast 
does  each  walk? 

44.  A  frontiersman  escaped  at  midnight  from  a  besieged 
post  and  started  at  4  miles  per  hour  for  a  fort  64  miles  away. 
He  was  met  by  a  cavalry  troop  which  had  started  from  the 
fort  at  3  A.M.,  going  9  miles  an  hour.  At  what  time  did  they 
meet? 

45.  A  trolley  car  started  at  noon  from  Boston  for  Wor- 
cester, a  44-mile  run.  At  12 :  30  a  carriage,  which  was  8  miles 
an  hour  slower  than  the  car,  started  from  Worcester  for 
Boston.   They  passed  at  2  p.m.    How  fast  did  the  carriage  go? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW        229 

46.  I  rode  6  miles  an  hour,  then  walked  7  times  as  long 
at  the  rate  of  4  miles  an  hour,  then  went  23  miles  by  train. 
I  traveled  91  miles  in  all.     How  long  did  I  walk? 

47.  A  man  starts,  on  a  wager,  to  walk  1000  miles  in  50 
days.  After  traveling  for  30  days,  he  finds  that  he  can  go 
a  distance  of  5  miles  less  per  day  and  still  have  time  to  go  20 
miles  extra  before  the  end  of  the  50  days.  How  many  miles 
per  day  did  he  walk  at  the  start? 

48.  A  village  block  680  feet  long  has  on  one  side  10  house 
lots.  Three  are  narrow  lots,  all  of  the  same  width;  the  other 
lots  are  20  feet  wider.     How  wide  are  the  lots? 

49.  A  wall  20  feet  high  is  built  with  8  courses  of  granite 
blocks  and  16  courses  of  brick.  Each  granite  block  is  22J 
inches  higher  than  the  thickness  of  a  brick.  How  thick  is 
each  brick? 

50.  A  platform  is  laid  with  boards  that  are  all  either  6 
inches  or  8  inches  wide.  There  are  5  more  of  the  8-inch 
boards  than  of  the  others,  and  the  platform  is  15  feet  wide. 
How  many  boards  were  used? 

51.  A  platform  8|  feet  wide  is  laid  with  43  hardwood 
strips,  of  which  16  are  each  an  inch  wider  than  each  of  the 
27  others.     What  are  the  widths  of  the  strips? 

52.  A  boy  starts  from  his  room  in  college  at  noon  one  day 
to  walk  to  his  home,  41  miles  away.  At  two  o'clock  of  the 
same  day,  the  father  starts  on  horseback  for  the  college, 
riding  3  miles  an  hour  faster  than  the  son  can  walk.  They 
meet  at  5  p.m.     How  fast  can  the  boy  walk? 

53.  A  pile  of  27  cannon  balls  in  three  sizes  weighs  254 
pounds.  There  are  13  of  the  smallest  size,  and  10  that  are 
twice  as  heavy.  The  others  are  each  by  2  pounds  the 
heaviest  in  the  pile.     Find  the  weight  of  each  size. 

54.  A  tank  holding  5940  gallons  is  filled  in  three  hours  by 
three  pipes.     The  first  pipe  carries  twice  as  much  as  tiie 


230    PROBLEMS  FOR  PRACTICE  AND  REVIEW 

third,  and  the  second  pipe  carries  3  gallons  less,  per  minute, 
than  the  third.  Find  the  number  of  gallons  per  minute 
passing  through  each  pipe. 

55.  A  reservoir  holding  10,080,000  gallons  is  filled  in  one 
day  by  three  pumping  stations.  The  first  station  delivers 
4  times  as  much  as  the  third,  and  the  second  station  delivers 
200  gallons  less,  per  minute,  than  the  third.  How  much 
does  each  deliver? 

56.  A  tank  holding  8820  gallons  is  drained  in  2  hours  by 
4  pipes,  marked  A,  B,  C,  and  D.  The  pipe  marked  A  car- 
ries one  gallon  per  minute  more  than  B,  and  2  gallons  per 
minute  less  than  C.  D  carries  half  as  much  as  all  the  other 
three.     What  is  the  capacity  of  each  pipe? 

57.  A  cistern  can  be  filled  from  the  water  main  in  12 
hours.  If  also  an  extra  pump  is  rigged  from  a  neighboring 
cistern,  it  can  be  filled  in  8  hours;  and  with  all  the  fire 
engines  in  town  to  help,  by  pumping  from  a  lake  near  by, 
it  could  be  filled  in  6  hours.  How  long  would  it  take  to  fill 
the  cistern  if  the  main  were  shut  off,  and  the  filling  depended 
entirely  on  the  extra  pump?  How  long  if  the  filling  depended 
entirely  on  the  fire  engines? 

58.  If  50  is  subtracted  from  \^  of  a  number,  the  result  is 
the  same  as  if  1  were  subtracted  from  j\  of  it.  Find  the 
number. 

59.  Subtracting  half  a  number  from  50  leaves  as  much  as 
subtracting  a  third  of  it  from  35.     Find  the  number. 

60.  Seven  ninths  of  a  certain  number  is  greater  than  25 
by  as  much  as  tV  of  it  is  less  than  6.     Find  the  number. 

61.  Two  thirds  of  a  certain  number  exceeds  99  by  as 
much  as  ^  of  it  is  less  than  31.     Find  the  number. 

62.  Five  sixths  of  a  certain  number  is  less  than  260  by  as 
much  as  f  of  it  exceeds  45.    What  is  the  number? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         231 

63.  Five  sixths  of  a  number  exceeds  27  by  as  much  as  tV 
of  it  is  less  than  6.     What  is  the  number? 

64.  If  a  number  is  added  successively  to  40  and  to  22, 
and  ^  of  the  first  sum  is  added  to  f  of  the  second  .sum,  the 
last  sum  will  be  28.     Find  the  number  added  to  40  and  22. 

65.  If  a  certain  number  is  subtracted  successively  from  40 
and  from  55,  and  then  ^  of  the  first  remainder  is  taken  from 
f  of  the  second,  the  last  remainder  will  be  10.  Find  the 
number  first  subtracted. 

66.  If  a  certain  number  is  subtracted  successively  from  53 
and  from  62,  and  then  f  of  the  first  remainder  is  taken  from 
y  of  the  second,  the  last  remainder  will  be  10.  Find  the 
number  first  subtracted. 

VI 

239.  Model  A.  —  At  what  time  between  4  and  5 
o'clock  are  the  hands  of  a  clock  9  minutes  apart? 

Let        X  =  the  number  of  minutes  past  4. 

Then    —^  =  the  number  of  minute  spaces  traversed  by  the  hour  hand 

in  X  minutes. 

And  20  +  ^2  =  the  number  of  minute  spaces  from  12  o'clock  to  the 

hour  hand. 
If  the  hour  hand  is  9  minutes  ahead  of  the  minute  hand,  the  equation 
becomes 

®     240  +  X  =  12  X  +  108  ®'  X  12 

®  132  =  11  a;  @  _  a;  _  108 

0  12  =  a;  ®  ^  11 

Ans.     12  minutes  past  4. 

Check:  At  4:12  o'clock,  the  hour  hand  is  ^  of  the  way  from  4  to  5, 
that  is,  at  the  21st  minute  mark;  the  minute  hand  is  at  the  12th  minute 
mark,  or  9  minute  spaces  behind. 


© 

20+^  +  9=: 

@ 

29+^2  =  . 

0 
® 

348  +   a;  =  12  a; 
348  =  11  a; 
31tV  =  X 

232         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

If  the  minute  hand  is  ahead  of  the  hour  hand,  the  equation  becomes 


®  same  values 

®  X12 

®  -a; 
®  -  11 
Ans.     28^  minutes  of  5. 

Of  these  two  answers,  the  first  only  can  be  realized  on  an  ordinary 
clock,  because  the  mechanism  of  the  clock  is  such  that  28x\  minutes  of  5 
is  never  indicated  by  the  position  of  the  hands.  They  move  by  jerks, 
one  for  each  tick  of  the  pendulum,  and  pendulums  that  tick  elevenths 
of  a  second  would  have  no  reason  for  existence  other  than  the  need  of 
illustrating  this  problem. 

Instead  of  the  numbers  4,  5,  and  9,  in  the  statement  of 
Model  B,  use  the  following  sets  of  numbers,  and  solve  the 
resulting  problems,  for  an  actual  clock: 


1. 

4,5,2 

6. 

5,  6,  14 

11. 

12,  1,  27 

16. 

12,  1,  22 

2. 

3,4,4 

7. 

6,  7,  14 

12. 

9,  10,  26 

17. 

4,  5,  24 

3. 

6,7,3 

8. 

8,9,7 

13. 

1,  2,  21 

18. 

7,  8,  24 

4. 

7,8,2 

9. 

8,9,4 

14. 

2,  3,  26 

19. 

9,  10,  23 

5. 

3,4,7 

10. 

5,6,3 

15. 

10,  11,  21 

20. 

1,  2,  17 

21.  A  student  of  music  always  practiced  at  the. piano  be- 
tween 5  P.M.  and  7  p.m.  One  day  he  noticed  as  he  began  to 
practice  that  the  hands  of  his  watch  were  exactly  3  minutes 
apart.  He  practiced  until  they  were  again  3  minutes  apart, 
and  then  stopped.     How  long  did  he  practice? 

22.  Find  when  the  hands  of  an  actual  clock  are  exactly 
together. 

23.  Find  when  the  hands  of  an  actual  clock  are  at  right 
angles. 

24.  When  are  the  hands  of  an  actual  clock  pointing  oppo- 
site ways? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW        233 

25.  When  are  the  hands  of  an  actual  clock  making  with 
each  other  an  angle  of  30°? 

26.  Find  when,  on  an  actual  clock,  the  hands  are  exactly 
one  minute  space  apart. 

VII 

1.  One  rhomboid  has  a  base  measuring  128  ft.  and  an 
altitude  of  56  ft.  Another  has  a  base  of  64  ft.  and  an  altitude 
of  112  ft.     What  is  the  ratio  of  their  areas? 

2.  One  triangle  has  an  altitude  of  1.732  in.  and  a  base  2 
in.  long.  Another  has  a  base  2  in.  long  and  an  altitude  of 
3.464  in.     What  is  the  ratio  of  their  areas? 

3.  One  rectangle  has  an  area  of  50  sq.'ft.  and  its  base  is  5 
ft.  long.  Another  has  an  area  of  125  sq.  ft.  and  its  base,  also, 
is  5  ft.     What  is  the  ratio  of  the  altitudes? 

4.  One  triangle  has  an  area  of  128  sq.  ft.  and  an  altitude 
of  16  ft.  Another  has  the  same  area  and  an  altitude  of  40  ft. 
What  is  the  ratio  of  the  bases? 

5.  One  rectangle  has  a  base  of  12.5  ft.  and  an  altitude  of 
10  ft.  Another  rectangle  has  a  base  of  100  ft.  and  an  altitude 
of  10  ft.     What  is  the  ratio  of  their  areas? 

6.  One  triangle  has  a  base  of  12  ft.  and  an  altitude  of  2i  ft. 
Another  triangle  with  the  same  base  has  an  altitude  of  36  ft. 
What  is  the  ratio  of  their  areas? 

7.  One  circle  has  a  radius  of  68  ft.  and  another  a  radius  of 
17  ft.     What  is  the  ratio  of  their  circumferences? 

8.  One  circle  has  a  radius  of  10  ft.  and  another  a  diameter 
of  125  ft.     What  is  the  ratio  of  their  circumferences? 

9.  Two  trapezoids  have  altitudes  of  125  ft.  and  1000  ft. 
One  trapezoid  has  bases  31  ft.  and  19  ft.  long;  the  other  has 
bases  24  ft.  and  36  ft.  long.    What  is  the  ratio  of  the  areas? 


234         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

10.  A  parallelogram  has  a  base  of  110  ft.  and  an  altitude  of 
44  ft.  Another  parallelogram  has  a  base  of  44  ft.  and  an 
altitude  of  110  ft.     What  is  the  ratio  of  the  areas? 

11.  One  rectangle  has  an  altitude  of  15.7  ft.  and  a 
base  of  52.9  ft.  Another  rectangle  has  a  base  of  75.6  ft. 
and  an  altitude  of  15.7  ft.  What  is  the  ratio  of  their 
areas? 

12.  One  triangle  has  a  base  1.732  in.  long  and  an  alti- 
tude 3.713  in.  long.  Another  has  a  base  of  1.732  in. 
and  an  altitude  of  2.902  in.  What  is  the  ratio  of  their 
areas? 

13.  One  rhomboid  has  a  base  196  ft.  long  and  an  altitude 
973  ft.  long.  Another  has  a  base  of  392  ft.  and  an  altitude 
of  973  ft.     What  is  the  ratio  of  their  areas? 

14.  One  rhomboid  has  a  base  of  4967  ft.  and  an  altitude 
of  8961  ft.  Another  has  a  base  of  59,670  ft.  and  an  altitude 
of  8961  ft.     What  is  the  ratio  of  their  areas? 

15.  One  trapezoid  has  bases  2.12  in.  and  3.56  in.  long,  and 
an  altitude  of  4.56  in.  Another  has  bases  2.72  in.  and  2.96 
in.  long,  and  an  altitude  of  7.42  in.  What  is  the  ratio  of  their 
areas? 

16.  One  trapezoid  has  bases  of  391.1  ft.  and  of  892.2  ft. 
and  an  altitude  of  10,910  ft.  Another  has  bases  409.6  ft. 
and  873.7  ft.  long  and  an  altitude  of  1910  ft.  What  is  the 
ratio  of  their  areas? 

17.  A  circle  has  a  radius  of  6  in.,  and  another  circle  has  a 
radius  of  1§  ft.     What  is  the  ratio  of  their  areas? 

18.  Two  circles  have  radii  of  2.76  ft.  and  5.02  ft.,  respec- 
tively.   What  is  the  ratio  of  their  areas? 

19.  Two  squares  have  perimeters  of  1.682  in.  and  3.084 
in.,  respectively.     What  is  the  ratio  of  their  areas? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW        235 

240.  Model  B.  —  The  perimeter  of  a  rhomboid  is  2.00 
in.  and  the  altitudes  are  1.03  in.  and  .387  in.     Find  the  area. 

S  S 

Since  S  =  api  =  bp2,  a  =—  and  6  =  —  and  a  +  b  =  half-perimeter. 

Pi  P2 

Whence,^3  +  -4  =  1.00. 

Solving  this  equation,  we  find  S  =  .281.     Ans. 

To  CHECK,  get  2  a  -\-  2  b  after  the  sides  have  been  found. 

Find  Sj  a,  and  b,  for  the  rhomboids  with  the  following  data: 

20.  pi  =  20,  p2  =  15,  perimeter  =  140 

21.  pi  =  12,  p2  =  9,  perimeter  =  56 

22.  pi  =  1.1,  p2  =  1-7,  perimeter  =  5.0 

23.  pi  =  15.9,  p2  =  18.1,  perimeter  =  157.6 

24.  pi  =  .371,  p2  =  .586,  perimeter  =  4.00 

25.  The  ratio  of  the  sides  of  a  rectangle  is  1.23,  and  the 
perimeter  is  78.5  in.     What  is  the  area? 

26.  One  base  of  a  rhomboid  exceeds  the  altitude  thereon 
by  twice  as  much  as  the  other  altitude  exceeds  the  other 
base.  The  longer  base  is  30.0  ft.  long  and  the  shorter  base 
is  18.3  ft.     Find  the  area. 

27.  The  altitudes  of  a  rhomboid  differ  by  3.50  in.  and  the 
bases  by  4.30  in.     The  perimeter  is  just  2  ft.     Find  the  area. 

28.  The  altitudes  of  a  rhomboid  have  a  ratio  1.05,  and 
they  differ  by  6  in.  If  the  shorter  base  is  100  ft.  long,  what 
is  the  area? 

29.  The  altitudes  of  a  rhomboid  have  a  ratio  2.13,  and  the 
perimeter  is  23.8  ft.     Find  the  area. 

30.  The  ratio  of  the  altitudes  in  a  rhomboid  is  2.30,  and 
the  ratio  of  one  altitude  to  the  base  it  stands  on  is  1.09.  The 
area  is  2.87  sq.  ft.     Find  the  perimeter. 

31.  The  ratio  of  two  sides  of  a  triangle  is  2.04,  and  the 
third  side  is  2.48  in.  less  than  their  sum.  The  perimeter  is 
1.86  in.     Find  the  sides. 


236         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

32.  Two  triangles  are  inscribed  in  the  same  stripe  2.46  ft. 
wide.  Their  bases  differ  by  7.62  ft.,  and  their  areas  have  a 
sum  of  39.37  sq.  ft.     Find  the  bases. 

33.  A  triangle  and  a  rhomboid  have  the  same  base,  12.37 
ft.  long;  the  altitudes  on  this  base  differ  by  8.23  ft.,  and  the 
sum  of  the  areas  is  215.6  sq.  ft.     Find  the  two  altitudes. 

34.  Two  triangles  inscribed  in  a  stripe  3.572  in.  wide  have 
bases  that  differ  by  2.873  in.,  and  the  ratio  of  their  areas  is 
11.57.     Find  the  bases  and  the  areas  of  the  triangles. 

35.  A  triangle  and  a  rhomboid  are  inscribed  in  the  same 
stripe.  The  base  of  the  triangle  is  35.8  in.  long  and  that  of 
the  rhomboid  is  13.7  in.  The  triangle  is  32.1  sq.  in.  larger 
than  the  rhomboid.     What  is  the  width  of  the  stripe? 

36.  Two  bases  of  a  triangle  are  respectively  65.4  ft.  and 
38.8  ft.  long,  and  the  corresponding  altitudes  differ  by  8.22 
ft.     Find  the  two  altitudes. 

37.  In  the  same  stripe  a  triangle  and  a  rhomboid  are 
inscribed,  with  the  base  of  the  rhomboid  bearing  to  the  base 
of  the  triangle  the  ratio  3.79.  The  rhomboid  is  635  sq.  in. 
larger  than  the  triangle.  The  stripe  is  20.6  in.  wide.  Find 
the  bases  of  the  triangle  and  the  rhomboid. 

38.  A  triangle  and  a  rectangle  have  bases  measuring  41.2 
cm.  and  53.7  cm.,  respectively.  The  triangle  bears  to  the 
rectangle  the  ratio  .327.  The  altitude  of  the  triangle  is  17.8 
cm.     What  is  the  altitude  of  the  rectangle? 

39.  Two  triangles  inscribed  in  the  same  stripe  have  bases 
15.7  in.  and  19.3  in.,  respectively,  and  the  difference  of  their 
areas  is  50.8  sq.  in.     Find  the  width  of  the  stripe. 

40.  The  altitude  of  a  trapezoid  is  3.74  ft.  The  bases 
differ  by  .0364  ft.  The  area-number  exceeds  the  sum  of  the 
length-numbers  of  the  bases  by  .0140.    Find  the  bases. 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         237 

41.  The  bases  of  a  trapezoid  are  1.032  in.  and  1.182  in. 
long.  The  area-number  exceeds  the  length-number  of  the 
altitude  by  1.09.     Find  the  altitude. 

42.  The  number  of  degrees  in  a  certain  arc  is  equal  to  the 
number  of  inches  in  its  complement.  If  each  degree  of  arc 
is  2  in.  long,  how  many  degrees  are  there  in  the  arc  first 
mentioned? 

43.  The  number  of  degrees  in  an  arc  exceeds  by  6  the  num- 
ber of  inches  in  its  complement;  and  the  entire  circumference 
is  75  ft.  long.  How  many  degrees  are  there  in  the  arc  first 
mentioned? 

44.  Two  supplementary  arcs  have  a  united  length  of  90  ft. 
The  number  of  inches  in  one,  added  to  the  number  of  degrees 
in  the  other,  gives  a  total  of  775.  How  many  degrees  are 
there  in  each  arc? 

45.  The  number  of  degrees  in  an  arc  is  equal  to  the  number 
of  inches  in  the  whole  circumference,  and  the  rest  of  the  cir- 
cumference measures  50  in.  What  central  angle  intercepts 
this  arc? 

46.  On  a  circle  308  in.  in  circumference,  the  number  of 
inches  in  a  certain  arc  is  equal  to  the  number  of  degrees  in  its 
complement.     How  many  degrees  are  there  in  the  arcs? 

47.  Of  the  two  angles  at  the  base  of  a  triangle,  one  is  3° 
less  than  double  the  other.  The  angle  at  the  vertex  bears 
to  its  own  supplement  the  ratio  If.  Find  the  angles  of  the 
triangle. 

48.  The  number  of  inches  in  an  arc  is  J  the  number  of  de- 
grees in  its  complement.  The  circumference  is  72  in.  Find 
the  length  of  the  arc  and  the  number  of  degrees  in  it. 

49.  The  altitude  of  a  parallelogram  is  3  in.  Its  base  is  7 
in.  longer  than  the  base  of  a  triangle  inscribed  in  the  same 
stripe.  The  areas  of  the  two  figures  have  a  ratio  3.4.  Find 
the  area  of  the  triangle. 


238        PROBLEMS  FOR  PRACTICE  AND  REVIEW 

50.  In  a  stripe  5.02  in.  wide,  a  triangle  and  a  parallelogram 
are  inscribed.  The  base  of  the  parallelogram  is  the  shorter 
by  3.37  in.,  and  the  ratio  of  its  area  to  the  area  of  the  triangle 
is  1.73.     Find  the  bases. 

51.  A  triangle  and  a  rhomboid  have  bases  respectively 
3.00  and  5.00  ft.  long.  The  altitude  of  the  triangle  is  the 
longer  by  14.52  in.,  and  its  area  bears  to  the  area  of  the  rhom- 
boid the  ratio  .500.  Find  the  two 
altitudes  and  the  area. 

A  sector  is  a  figure  bounded  by  an 
arc  of  a  circle  and  by  the  sides  of  the 
central  angle  that  intercepts  the  arc.  If 
the  central  angle  is  A  degrees,  the  area 

of  the  sector  will  be  ^^7:  of  the  area  of 

the  entire  circle. 

52.  Find  the  area  of  the  sector  of  a  circle  of  radius  20  in., 
if  the  central  angle  is  75°. 

53.  Find  the  area  of  a  circle,  if  a  sector  of  38°  has  an  area 
of  43.8  sq.  in. 

64.  Find  the  area  of  the  sector  of  a  circle  of  radius  32.7 
cm.,  if  the  central  angle  is  83.7°. 

55.  In  a  circle  whose  radius  is  3.37  in.,  how  wide  a  central 
angle  will  contain  a  sector  of  3.37  sq.  in.? 

56.  In  a  circle  of  28.7  ft.  radius,  a  sector  has  a  total  per- 
imeter of  83.3  ft.     How  many  degrees  are  there  in  the  sector? 

57.  In  a  circle  of  radius  30.8  in.,  a  sector  contains  283  sq. 
in.     How  many  degrees  are  there  in  it? 

58.  Two  trapezoids  inscribed  in  the  same  stripe  have  one 
base  the  same;  the  other  bases  are  f  of  an  inch  and  |  of  an 
inch  respectively.  The  ratio  of  their  areas  is  f .  Find  the 
comimon  base. 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         239 

59.  The  areas  of  two  circles  differ  by  92.8  sq.  in.  The 
area  of  a  sector  of  40°  on  one  circle  exceeds  the  area  of  a  sec- 
tor of  60°  on  the  other  by  8.35  sq.  in.  Find  the  areas  of  the 
circles. 

60.  The  base  angles  of  a  triangle  differ  by  one  degree. 
The  sum  of  these  angles  exceeds  the  angle  at  the  vertex  by  as 
much  as  half  the  larger  base  angle.  Find  the  angles  of  the 
triangle. 

61.  In  two  circles,  one  circumference  is  double  the  other. 
An  arc  of  the  smaller  circle  exceeds  its  complementary  arc  on 
the  larger  circle  by  tV  of  the  smaller  circumference.  How 
many  degrees  are  there  in  each  arc? 

62.  The  point  C  cuts  off  from  the  quadrant  AB  an  arc, 
AC,  of  38°.  How  far  shall  the  point  C  be  moved  in  order 
that  the  ratio  of  AC  to  CB  shall  be  GJ? 

63.  The  bases  of  two  rhomboids  differ  by  2f  ft.  Their 
altitudes  are  17.3  ft.  for  the  one  with  the  shorter  base,  and 
23.8  ft.  for  the  other.  The  ratio  of  their  areas  is  |.  Find 
the  areas. 

64.  A  certain  circle  has  twice  the  area  of  a  certain  rectangle. 
If  the  rectangle  were  125  sq.  ft.  smaller,  the  circle  would  be 
3  times  as  large  as  the  rectangle.     What  are  the  areas? 

65.  A  rectangle  is  380.1  sq.  ft.  larger  in  area  than  a  tri- 
angle. If  the  triangle  were  9.52  times  as  large,  the  two  areas 
would  be  the  same.     What  are  the  areas? 

66.  One  triangle  is  4  times  as  large  as  another,  and  the 
sum  of  fhe  areas  of  both  is  one  half  the  area  of  a  certain  circle. 
If  the  larger  triangle  were  5  times  as  large,  and  the  smaller 
triangle  were  100  sq.  ft.  less,  the  area  of  the  circle  would  be 

•just  equal  to  the  area  of  the  two  triangles.    What  are  the 
three  areas? 

6V.   What  is  the  complement  of  an  angle  of  47.15°? 


240         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

68.  What  is  the  supplement  of  an  angle  of  47.15°?  What 
is  the  supplement  of  the  complementary  angle? 

69.  Whatis  the  complement  of  an  angle  of  89.175°?  What 
is  the  supplement  of  the  same  angle?  What  is  the  supple- 
ment of  the  complementary  angle? 

70.  What  is  the  supplement  of  an  angle  of  126.52°?  What 
is  the  complement  of  the  supplementary  angle? 

71.  What  is  the  complement  of  an  arc  of  62.33°?  What 
is  the  supplement  of  the  same  arc? 

72.  Two  complementary  arcs  have  a  ratio  4.  What  are 
they? 

73.  Two  supplementary  arcs  have  a  ratio  6J.  What  are 
they? 

74.  Of  two  supplementary  arcs,  one  is  6.89  times  the  other. 
What  are  the  arcs? 

75.  The  ratio  of  two  complementary  arcs  is  1.232.  What 
are  the  arcs? 

76.  Two  angles  of  a  triangle  are  56.32°  and  72.49°.  What 
is  the  third  angle? 

77.  The  sum  of  tw^o  angles  of  a  triangle  is  171.2°.  What 
is  the  third  angle? 

78.  One  angle  of  a  triangle  is  36.2°.  Of  the  other  two 
angles,  one  is  twice  the  other.  What  are  the  other  two 
angles? 

79.  One  angle  of  a  triangle  equals  half  the  largest  and 
twice  the  smallest  angle  of  the  triangle.  What  are  the  angles 
of  the  triangle? 

80.  One  angle  of  a  triangle  is  91.3°.  The  other  two  angles 
are  equal.    What  are  they? 

81.  In  a  triangle  with  two  equal  angles,  the  ratio  of  the 
third  angle  to  the  sum  of  the  other  two  is  1.672.  What  are 
the  angles  of  the  triangle? 


PROBLEMS  FOR  PRACTICE  AND  REVIEW    241 

82.  In  a  triangle  with  two  equal  angles,  the  ratio  of  each 
of  the  equal  angles  to  the  third  angle  is  2.179.  What  are  the 
angles  of  the  triangle? 

83.  In  a  triangle,  the  ratio  of  the  largest  angle  to  the 
smallest  is  5.234,  and  the  ratio  of  the  third  angle  to  the 
smallest  is  1.212.     What  are  the  three  angles? 

84.  One  of  two  supplementary  arcs  of  a  circle  of  16.3  in. 
radius  is  2.7  in.  longer  than  the  other.  Find  the  number  of 
degrees  in  each  arc. 

VIII 

241.  In  multiplying  and  dividing  algebraic  polynomials,  it  is 
of  great  importance  to  arrange  not  only  the  given  expressions 
but  all  the  expressions  obtained  in  the  course  of  the  work 
according  to  the  powers  of  some  one  letter.  For  example, 
the  expression 

Qa'h  +  Q  ah^  +  a«  +  6'  +  15  a^b^ 
should  be  rearranged  in  one  of  the  two  following  ways: 

a6  +  6  w^h  +  15  a^62  +  6  ab^  +  h^ 
or  6^  +  6  afes  +  15  a'W  +  6  a^b  +  a« 

Rearrange  the  following  expressions,  if  necessary,  before 
multiplying: 

1.  (x  +  9)  (x2  +  4a;  -45) 

2.  {x  +  9)  (x2  -  36  -  5  a;) 

3.  (9  +  a;)  (27  +  6  re  -  x") 

4.  (x  +  9)  (72  -  x^  +  x) 

5.  (3  a:  +  a;2  -  10)  {x  -  5) 

6.  {p'-q'  +  7vq)(7p-q) 

7.  {x^  -  14  -  5  x)  (2  -  a;) 

8.  (x2  -  14  -  5  a;)  (x  +  7) 


242         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

0.  (4  A;  +  h)  (/i2  -  4  /c2  -  3  hk) 

10.  {d'  -{-g'-  dg)  id'  +  g'  +  dg) 

11.  (10  2/2  -  3  X2/  +  x')  (x  +  3  2/) 
^^12.  (10  2/2  +  a;2  -  3  xy)  (x  +  5  2/) 

13.  (10  2/2  +  a;2  -  3  xy)  {x-\-2y) 

14.  (c2  +  6  cd  -  5  d")  (6  d  -  c) 

15.  (c2  +  6  c(^  +  5  ^2)  {bd-  c) 

16.  (14  y'  -]-  5xy  —  x')  (x  -\-  7  y) 

17.  (14  2/2-^2  +  5  0:2/)  (x  -2y) 

18.  (5  a2  +  5  a;2  _|_  2  aa;)  (a^  +  x'  -  5  ax) 

19.  (a;2  +  4  +  2  x)  (a;2  +  4  -  2  x) 

20.  (x2  -  15  +  2  x)  (2  X  +  15  -  x2) 

21.  (r^  —  r's  +  ^s^  —  s^)  (r^  +  rs  +  s^) 

22.  (2  x2  +  9  2/2  +  12  X2/)  (12  X2/  -  2  x2  -  9  y') 

23.  (x2  +  6  X  +  18)  (18  +  x2  -  6  x) 

24.  (10  p5  +  2  p2  +  25  g2)  (10  pg  -  2  p2  _  25  g^) 

25.  (3  /i2  +  4  s2  +  2  hs)  (2hs-Sh'-4:  s^) 

26.  (x2  +  3  X2/  +  2  2/2)  (x2  -  6  X2/  +  4  2/^) 

27.  (3  a2  +  3  52  -  10  ah)  (Sa'  -Sh'  -S  ah) 

28.  (4  x2  -  9  2/2)  (3  2/2  -  5  X2/  -  2  x2) 

29.  {p^  +  2pq  +  Sq){Sp  +  2q)        4 

30.  (5  a2  -  3  ax  +  2  a^)  {a'  +  2  ax) 

31.  (a3  +  53  +  3  ^25  _^  3  ^52)  (3  ^5  _  ^2  _  2  62) 

32.  (/i^  +  /b^  +  6  /i2/c2  -  4  /i/b  [/i2  +  /c2])  (4  /i/c  +  /i2  +  h^) 

33.  (5  2/2  -  2  X2/  +  3  x2)  (2  X2/  +  5  2/2  +  3  x2) 

34.  (2  X2/  -  5  2/2  -  3  x2)  (4  X2/  +  10  2/2  +  6  x2) 

35.  (3  x2  +  3  X2/  +  2  2/2)  (2  2/2  -  3  X2/  +  3  x2) 

36.  [5-p3  +  p(2_3p)][l-3p(p-3)] 

37.  (x2  +  fx  +  i)(f  -ix  +  x2) 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         243 

38.  (3  a2  +  ^  -  2  a)  (5  a2  -  i  -  i  a) 

39.  (x2  +  xy  +  i  y^)  (f  y^  -  xy  +  x^) 

40.  (ix^-ix-\-^){ix  +  ^x^  -i) 
41.-  (a^  +  ¥)  {a  -  h)  (a^  +  a6  +  ¥) 

42.  (a2  +  2  a6  +  62)  (a^  -  2  ah  +  ¥) 

43.  (a^  +  a%  +  a62  +  6^)  (a  -  b) 

44.  (a2  +  62)  (a2  -  62)  -  (a  -  6)^ 

45.  (a3+3  a26+3  a62+63)  (a+6)  -  (6-a)  (a3+a26+a62+63) 

Rearrange  the  following  expressions,  if  necessary,  before 
dividing: 

46.  (x3  -  9  x2  +  8  a;  -  12)  -^  (x2  +  7  x  -  6) 

47.  (7  fc3  +  9  /c2  +  14  A;  +  12)  -^  (7  fc2  +  2  A;  +  12) 

48.  (7  x3  -  12  x2  -  8)  -^  (7  x2  +  2  a;  +  4) 

49.  (2  a^  -  a62  +  6^)  --  (2  a2  -  2  a6  +  62) 

50.  (61/3  +  6  X2/2-34  X22/-4  x^)  -t-  -  (6  2/2+18  x?/  +  2  x^) 

51.  (x3-  2x  +  1)  ^  (x-  1) 

52.  (2  a3  +  a26  -  13  a62  +  28  6^)  -v-  (2  a  +  7  6) 
''^.  (12  x3  +  8  ^22/  -  21  x?/2  +  14  2/3)  --  (3  X  +  2  ?/) 

54.  (2  xV  -  11  0^22/2  -  57xy  +  31)  -^  {2  xy  -  1) 

55.  (14  d3  +  15  d2g,  -  58  dg^  +  21  g^)  -^  {7  d  -  3  g) 

56.  (a'  +  4)  -^  (a2  +  2  a  +  2) 

57.  (4  2/^  +  81)  ^  (2  2/2 +  6  2/ +  9) 

58.  (d^  +  324)  ^  (^2  +  6  c^  +  18) 

59.  (64  p'  +  81)  -^  (8  p2  _  12  p  +  9) 

60.  (6^  +  962  +  81)  -^  (62  +  36  +  9) 

61.  (n'  +  n2/b*  +  k^)  -^  (^2  +  nk^  +  fc^) 

62.  (p4  -  27  p2g2  _|.  q4)   _^   (p2  _  5  p^  _  ^2) 

63.  (81  xy  +  8  xY  +  16)  -^  (9  x22/2  -  8  X2/  +  4) 

64.  (a«6«  +  14  a^64  +  625)  -^  (a^ft*  +  6  a%^  +  25) 


244         PROBLEMS  FOR  PRACTICE  AND  REVIEW 

65.  (a2  -  ¥  -  2hc  -  c^)  -^  {a  -  h  -  c) 

66.  (x^  +  x^y  +  xY  +  xY  +  xy*  +  y^)  -i-  (x^  +  y^) 

67.  (8  c^-22  c3a;+43  c^o^^-SS  0x^+24 x^)  -^ (2  c^-S  cx+4  a;2) 

68.  (a^  +  63  +  c^  -  3  a6c)  -f-  (a  +  6  +  c) 

69.  (s^  -  2  s3  +  1)  ^  (§2  -  2  s  +  1) 

Vo.  (x^  -  oi^a^  -  x2  +  a2)  -^  -  (x2  -  xa  -  X  +  a) 

71.  (r®  —  r^s^  —  r^  +  s^)  -^  (r^  +  y*^  —  ^'^s  +  r  —  rs  —  s) 

72.  (/i^o  +  /iV'  +  g'')  -  (/i'  +  /ig'  +  Sf=^) 

73.  (la:3  +  2)-^(ix+l) 

74.  (|a3-ia62-^63)  ^(3^  +  5) 

75.  (I  a^  +  I  a63  -  I  64)  --  (a  +  1 6) 

76.  (i  rc3  _|_  2  ^5^2  _  ^252^  _  I  ^353)  ^  (1  a,  +  ^6) 

77.  aa^  +  ia262+|64)^Q62  +  ict6  +  ^a2) 

Wherever  there  is  a  remainder  in  the  following  divisions, 
write  the  remainder  as  the  numerator  and  the  divisor  as  the 
denominator  of  a  fraction  which  is  an  additional  term  of  the 
complete  quotient: 

78.  (x3  +  2/3)  -^  (x-  y) 

79.  (x^  _  6  ^2  _^  12  X  -  6)  -f-  (x  -  2) 

80.  (a^  +  65)  -^  (a2  +  4  a  +  8) 

81.  (a  +  a^  -  34  a2  +  226)  -^  (a^  -  25) 

82.  (200  +  3  a  -  34  a2  +  a^)  -^  (a^  -  4) 

83.  (300  +  a^  -  2  a  -  34  a2)  -^  (a2  _  2  a  -  15) 

84.  (150  +  a*-34a2) -^  (a3  +  20-a(5a  +  4)) 

85.  (a3  +  a26+^a62  +  |)^(|-a) 

86.  (xio  +  xy  +  2/'")  -^  (x^  +  2/') 

87.  [{x^  -  y^y  +  x!/  (x2  +  2/2  +  3  xi/)]  -^  (x?  +  xy  +  y^) 


PROBLEMS  FOR  PRACTICE  AND  REVIEW    245 

IX 

Plot  the  loci  of  the  following  equations: 

1.  2  x2  -  5  X  +  3  2/  =  171        11.  xy  =  36 

2.  3 x2  +  3 X  -  13 2/  =  239      12.   x"" -\- 2Q  =  3xy  +  2y 

3.  y^-4:xy-\-Sy  =  lQx-7-4:X^  13.   5x^-{-Qxy  =  144 

4.  x2  +  2  X2/  -  24  14.   x2  +  1/2  +  6  X  -  4  2/  =  12 

5.  x2  +  x?/  +  2  2/2  =  74  15.   X2/  +  3  X  -  4  2/  =  36 

6.  24  2/2  -  25  X2/  =  25  x^  le.   9x^+42/^+145  =54x  -402/ 

7.  x2  +  2/2=5  17.  9x2-55  =42/2+54x+402/ 

8.  2  x2  +  X2/  +  2/2  =  8  18.   2/^  -  8  X  +  28  =  4  2/ 

9.  x2  -  X2/  +  2/2  +  3  X  =  34     19.   x2  =  8  2/ 

10.  x2  +  2/2  =  53  20.   x2  +  25  =  6  X  +  8  2/ 

21.  Find  where  each  of  the  loci  of  the  equations  in  Exs. 
1-20  intersects  the  line  x  +  2  2/  =  5. 

X 

Solve  the  following  pairs  of  simultaneous  equations: 

I.  x-22/=9  e.   x-2y  =  6.42 

X2/  +  X  =  2/  +  9.98 

7.  2  X  +  2/  =  8.45 
a;2  _  2/2  =  6.18 

8.  2  X  +  2/  =  9.45 
x2  +  22/=2/'  +  2x  +  6.18 

9.  140  x2  -  30  X2/  =  23 
5  X  +  2/  =  2.7 

10.   7  x2  -  6  X2/  =  40.6 
5  X  +  2  2/  =  9.85 

II.  In  two  triangles,  the  vertical  angles  are  equal,  and 
the  ratio  of  the  right  sides  exceeds  by  1  the  ratio  of  the  left 
sides.  The  ratio  of  the  areas  of  the  triangles  is  \^.  What 
are  the  ratios  of  the  sides? 


xy  =  68 

2. 

X  -  2  2/  =  472 

xy  =  29952 

3. 

X  -  2  2/  =  3.42 

xy  =  8.98 

4. 

2  X  +  2/  =  13 

x2  -  2/2  =  16 

5. 

2  X  +  2/  =  755 

a;2  -  2/2  =  6487 

246    PROBLEMS  FOR  PRACTICE  AND  REVIEW 

12.  In  two  triangles,  the  vertical  angles  are  equal,  and 
the  ratio  of  the  right  sides  is  f  the  ratio  of  the  left  sides. 
The  ratio  of  the  areas  is  |.     What  are  the  ratios  of  the  sides? 

13.  Two  triangles  have  vertical  angles  equal,  and  the 
ratio  9f  the  left  sides  is  90.21  per  cent  of  the  ratio  of  the  right 
sides.  The  ratio  of  the  areas  is  1.216.  What  are  the  ratios 
of  the  sides? 

14.  Two  sides  of  a  triangle  are  28  cm.  and  27  cm.  Half 
the  area  is  cut  off  by  a  line  that  cuts  the  longer  side  3  cm. 
higher  up  than  the  point  where  it  cuts  the  shorter  side. 
Where  does  the  line  cut  the  shorter  side? 

15.  The  ratio  of  the  areas  of  two  similar  triangles  exceeds 
the  ratio  of  the  bases  by  42.  If  the  area  of  the  smaller  tri- 
angle is  40  sq.  in.,  what  is  the  area  of  the  larger? 

16.  In  a  trapezoid,  the  upper  and  lower  bases  are  56  cm. 
and  70  cm.,  and  the  right  side  is  15  cm.  If  the  sides  are 
extended  until  they  meet,  how  far  up  will  the  left  side  cut  the 
right? 

17.  The  ratio  of  the  areas  of  two  similar  triangles  is  10 
less  than  .763  times  the  ratio  of  the  bases.  If  the  area  of  the 
larger  triangle  is  82.31  sq.  ft.,  what  is  the  area  of  the  smaller? 

18.  The  ratio  of  the  areas  of  two  similar  triangles  is  12 
less  than  7  times  the  ratio  of  the  bases.  If  the  area  of  one 
triangle  is  5  sq.  mi.,  what  is  the  area  of  the  other?  (Four 
answers.) 

19.  The  ratio  of  two  rectangles  is  If  greater  than  the 
ratio  of  their  bases,  and  3  less  than  the  ratio  of  their  alti- 
tudes.    What  is  the  ratio  of  the  rectangles? 

20.  In  two  rectangles  which  have  equal  areas,  the  ratio 
of  the  bases  exceeds  by  |  the  ratio  of  the  altitudes.  What 
is  the  ratio  of  the  rectangles? 

21.  The  ratio  of  the  bases  of  two  rectangles  exceeds  by  1 
the  ratio  of  the  altitudes;  and  the  sum  of  these  two  ratios  is 


PROBLEMS  FOR  PRACTICE  AND  REVIEW         247 

1  less  than  the  ratio  of  the  rectangles.     What  is  the  ratio  of 
the  rectangles? 

22.  The  altitude  of  a  triangle  is  25  per  cent  greater  than 
the  base.  Its  area  is  equal  to  a  rectangle  having  an  altitude 
of  9  cm.  and  a  base  2  cm.  shorter  than  the  base  of  the 
triangle.     Find  the  altitude  and  the  base  of  the  triangle. 

23.  One  side  of  a  triangle  is  11  cm.  longer  than  the  other. 
A  line  cutting  these  sides,  at  distances  of  6  cm.  and  7  cm., 
respectively,  from  the  vertex,  cuts  off  a  quadrilateral  equal  to 
f  the  original  triangle.     Find  the  lengths  of  the  sides. 

24.  The  two  sides  of  a  right  triangle  differ  by  |  the  hypot- 
enuse, and  the  hypotenuse  is  f  their  sum.  Find  the  three 
sides. 

The  slant  side  of  a  right  triangle  is  called  the  hypotenuse. 

25.  The  two  sides  of  a  right  triangle  differ  by  2  cm.,  and 
the  hypotenuse  is  8  cm.  long.    Find  the  length  of  the  two  sides. 

26.  One  side  of  a  right  triangle  is  12  cm.  more  than 
double  the  other,  and  the  hypotenuse  is  78  cm.  long.  Find 
the  lengths  of  the  legs. 

27.  One  side  of  a  right  triangle  is  1  cxa.  less  than  4  times 
another,  and  the  hypotenuse  is  14  cm.  less  than  the  sum  of 
these  two  sides.     Find  the  three  sides. 

28.  The  hypotenuse  of  a  right  triangle  is  11  cm.  greater, 
and  the  larger  leg  is  2  cm.  greater,  than  double  the  shortest 
side.     Find  the  three  sides. 

29.  In  a  triangle  with  two  equal  sides,  the  area  is  120  sq. 
cm.  One  of  the  equal  sides  is  17  cm.  Find  the  base  and 
the  altitude. 

30.  The  length  of  a  rectangle  is  J  centimeter  less  than  5 
times  the  width,  and  the  diagonal  is  44  millimeters.    Find 

the  area. 

I  meter  =  100  centimeters 
1  centimeter  =  10  millimeters 


248    PROBLEMS  FOR  PRACTICE  AND  REVIEW 

31.  One  side  of  a  right  triangle  is  |  meter  less  than  the 
hypotenuse,  and  the  whole  perimeter  is  66  centimeters  less 
than  3  times  the  hypotenuse.     Find  the  sides. 

32.  Of  the  numbers  representing  the  three  angles  of  a 
triangle,  two  are  formed  by  the  same  digits  interchanged, 
and  the  other  is  the  square  of  the  sum  of  those,  digits.  The 
first  two  angles  would  be  complements  if  one  were  diminished 
by  9°.     What  are  the  angles? 

33.  The  area  of  a  rectangle  is  480  square  centimeters,  and 
the  diagonal  is  34  centimeters.     Find  the  perimeter. 

34.  The  length  of  a  rectangle  exceeds  the  width  by  49 
miUimeters,  and  the  perimeter  exceeds  the  sum  of  the  diago- 
nals by  2  centimeters.     Find  the  sides. 

6  35.  The  diagonal  of  a  rectangle  exceeds  5  times  the  width 
by  2  centimeters,  and  the  perimeter  is  1Q4  millimeters. 
Find  the  sides. 


INDEX 


Abbreviating  explanations  of  prob-      Axiom,  41 . 


lems,  11. 
Ability  and  time,  problem  of,  100. 
Accm-acy,  18. 

limited  by  data,  59. 

of  diagrams,  171. 
Acre  =  160  square  rods,  206. 
=  .4047  hectares,  60. 

2.471  acres  in  a  hectare, 
60. 
Addition,  approximate,  59. 
Age  problems,  38. 
Air,  weight  of,  20,  60. 
Algebra,  11. 

ancient  textbooks,  42. 

derivation  of  the  word,  42. 
Algebraic  expression,  13. 
Algebraic  negative,  136. 
Alphabet,  use  of,  in  algebra,  23. 
Altitude,  67. 
Analjdiic  geometry,  214. 
Angle,  sides  of,  22. 
Angles,  measurement  of,  21. 

of  a  polygon,  164. 

in  a  stripe,  32. 
Answers,  complete  Ust  of,  175. 

meaningless,  140. 
Approximate  addition,  59. 

computation,  50. 

division,  54. 
Arc,  47. 
Area,  ratios  of,  148. 

of  circle,  78. 

of  trapezoid,  75. 

of  triangle,  71. 


Axiom  A,  130. 

Axis  of  X,  170. 

of  y,  170. 

Base,  67. 
Binomial,  111. 
Brace,  108. 
Bracket,  108. 

Center,  46. 

Centigrade  and  Fahrenheit,  175. 

Change  of  form,  42. 

Changes  of  value  in  an  equation, 

41,  83. 
Check,  by  constructing  loci,  186. 

for  approximate  division,  55. 

for  approximate  multipUcation, 
53. 

for  long  division,  117. 
Checking,  14. 

by  coefficients,  106,  111. 

quadratics   with   the   standard 
parabola,  215. 
Circle,  46,  78. 

equation  of,  208. 
Circumference,  46,  78. 

length  of  the,  78. 
Clock  problems,  231. 
Coal,  components  of,  20. 

weight  of,  53. 
Coefficient,  41. 

of  a  term,  41. 
Coefficients,  check  by,  106. 
Complementary  arcs,  47. 


249 


250 


INDEX 


Complements,  23. 

Complete  quotient,  244. 

Completing  the  square,  127. 

Composition  of  motion,  95. 

Computation,  approximate,  50. 

Condition,  equation  of,  119. 

Conditions,  to  determine  two  un- 
knowns, 181. 

Continued  multiplication,  109. 

Convention,  14. 

Converse  statements,  179. 

Cord  (of  wood)  =4X4X8  feet, 
136. 

Correspondence    of    angles    and 
sides,  155. 
of  numbers,  169. 

Cross  products,  122. 

Cube  of  a  number,  110. 

Cube  root,  110. 

Cubic  meter  =  1.3079  cubic  yards, 
60. 

Cubic  yard   (1.3079  cubic  yards 
in  a  cubic  meter),  60. 

Current  problems,  98. 

Curve  of  a  railroad  track,  61. 

Cylinder  (lateral  surface),  133. 

Data,  accuracy  of,  18,  58. 
Decimal  data,  18. 

point,  placing  of,  50,  51. 

subdivision,  18. 
Degree,  21. 

(57.296°  in  a  radian),  60. 

of  a  term,  110. 

of  an  expression,  110. 
Diagram,  algebraic,  170. 
Diameter,  47. 
Difference  of  squares,  120. 

of  unknown  numbers  given,  30. 
Digits,  problem  of  the,  92. 
Discarding  figures,  51. 


Distributive  factoring,  113. 

law,  112. 
Division,  approximate,  54. 
Doubtful  columns,  52,  54. 

figures,  51. 

Egyptians,  147. 
Eight-figure  accuracy,  18. 
Elimination  by  combination,  186. 

by  substitution,  203. 
Ellipse,  214. 
Equation,  13. 

linear,  200. 

of  condition,  119. 

quadratic,  121,  130,  200. 
Equations,  construction  of,  91. 

inconsistent,  198. 

independent,  198. 

indeterminate,  199. 

simultaneous,  181. 
Equiangular  polygon,  165. 

triangle  is  equilateral,  166. 
Equilateral  polygon,  165. 

triangle  is  equiangular,  166. 
Equivalents  of  inches  in  tenths  of 

a  foot,  53. 
Errors  of  rejection,  52,  53. 
Explanations  of  problems  abbre- 
viated, 11. 
Explicit  formulas,  80. 
Exponent,  110. 

Factoring,  by  cross  multipUcation, 
123. 

distributive,  113. 
Factors,  109. 

of  an  equation,  131. 
Fahrenheit  and  Centigrade,  175. 
Figures  inscribed  in  a  stripe,  67. 
Five-figure  accuracy,  18. 
Fixing  the  decimal  point,  50,  51. 


INDEX 


251 


Foot  (5280  feet  in  a  mile),  190. 
Formula,  63. 

S  =  ab,  63. 

S  =  api  =  hp2,  68. 

S  =  hapi  =  hbp2  =  h cps,  71. 

5  =  1  p  (6,  +  62),  75,  80. 

c  =  2irr,  5  =7rr2,  78. 

s   =  vt,  95. 

TF  =  (n  -  2)  180^  164. 
Fractional  equations,  83. 

Gallon  =  3.785  liters,  60. 
Geometry,  147. 
Greeks,  147. 

Hectare  =  2.471  acres,  60. 

.4047  hectares  in  an  acre,  60. 
History  of  equations,  14. 
Hyperbola,  214. 

Identical  equation,  119. 

Identities,  119. 

Imaginary  roots,  144. 

Implicit  formulas,  80. 

Inches  reduced  to  tenths  of  a  foot, 

53. 
Inconsistent  equations,  198. 
Independent  equations,  198. 
Indeterminate  equations,  199. 
Index,  ilO. 

Indirect  measurement,  65. 
InequaUty,  sign  for,  72. 
Inscribed,  figures,  in  a  stripe,  67. 
Integer,  17. 
Intercepts,  178. 
Irrational  numbers,  142. 
Italian  method  of  division,  56. 
of  subtraction,  56. 

Kilogram  =  2.20  pounds,  20. 
.4536  kilograms  in  a  pound,  60. 


Law  of  signs,  115,  116. 
Left-hand  order  of  partial  prod- 
ucts, 50. 
Length,  measurement  of,  17. 

of  the  circumference,  78. 
Limits  of  accuracy,  18. 
Linear  equation,  200. 
Linear-quadratic  pairs,  200. 
Liter  (3.785  Uters  in  a  gallon),  60. 
Locus  of  an  equation,  175. 

of  a  two-letter  quadratic,  214. 
Long  division,  117. 

Measurement,  indirect,  65. 

of  angles,  21. 

of  length,  17. 
Measurement-numbers,  50. 
Member  of  an  equation,  13. 
Meter,  17, 

=  39.370  inches,  20,  56. 
Mile  =  5280  feet,  190. 
Minus  terms,  104. 
Minute  (60  minutes  in  a  degree), 

22. 
Monomial,  111. 
Motion,  composition  of,  95. 
Multiplication,  continued,  109. 

(distributive  law),  36,  37. 

how  indicated,  14. 

of  polynomials,  114. 

reversed,  50. 

Natural  order  of  partial  products, 

50. 
Negative    answers,    meaning    of, 

136,  138. 
Negative  numbers  (subtrahends), 

103. 
Negative  terms  (subtrahends),  37, 

104. 
Notation  in  geometry,  147. 
NuU  root,  144. 


252 


INDEX 


Number-pair,  208. 
Number  TT,  78. 
Numbers,  irrational,  142. 

Odd  and  even  angles  in  a  stripe, 
32. 

Pair  of  straight  lines,  214. 
Parabola,  214. 
Parallel  lines,  32. 
Parallelogram,  68. 
Parentheses,  removal  of,  89,  106. 

within  parentheses,  108. 
Partial    products,    as    successive 
corrections,  51. 

in  division,  54. 

order  of,  50. 
Perimeter,  235. 
TT,  78. 
Plotting,  of  points,  171. 

a  quadratic,  207. 
Plus  terms,  104. 
Polygon,  angles  of  a,  164. 

equiangular,  equilateral,  165. 

regular,  166. 
Polynomial,  111. 

Polynomials,     multiplication     of, 
114. 

rearrangement  of,  241. 
Positive  terms,  104. 
Pound  =  .4536  kilograms,  60. 
Power,  110. 
Precision  of  measurement,  18. 

of  instruments,  50. 

of  data,  58. 
Problem    of    ability    and    time, 
100. 

of  the  clock,  231. 

of  the  digits,  92. 

of  two  velocities,  94. 
Protractor,  23. 


Pure  quadratic,  143. 
Pythagorean  theorem,  162. 

Quadrant    (quarter   of    a   circle), 

23. 
Quadratic  equation,  121,  130,  200. 
Quadratic  expression,  121. 
Quadratic  loci,  210. 
Quadratic  products,  121. 
Quadratics,  peculiar,  144. 

Radian  =  57.296^  60. 
Radius,  47. 
Railroad  curve,  61. 
Ratio,  16. 

Ratios  obtained  by  measuring  an- 
gles, 65. 
Ratios  of  area,  148. 
Reciprocals,  solving  for,  194. 
Rectangle,  area  of,  63. 
Reducing  an  equation,  83. 

an  expression,  83. 
Regular  polygon,  166. 
Rejecting  figures,  51,  62. 
Rejection,  errors  of,  52. 
Rejection  of,  5,  53. 
Removal  of  parentheses,  89,  106. 
Rhomboid,  area  of,  67. 
Right  triangles,  similar,  159. 
Rod  =  16.5  feet, 

(320  rods  in  a  mile). 
Root,  110. 
Roots,  equal  and  opposite,  143. 

of  an  equation,  131,  143. 
Rough  check,  51. 
Rule  for  solving  simple  equations, 
42. 

for  subtraction,  107. 

Scale,  algebraic,  168. 
of  a  diagram,  172. 


INDEX 


253 


Second  (60  seconds  in  a  minute), 

22. 
Sector,  238. 

Shop  method  of  subtraction,  56. 
Shortages,  37. 
Sign  of  identity,  119. 

of  perpendicularity,  177. 

of  similarity,  163. 
Signs,  law  of,  115,  116. 
Similar  figures,  155.  ^ 

polygons,  155. 

right  triangles,  159. 

terms,  41,  111. 

triangles,  156. 
Simple  equation,  study  of,  41. 
Simultaneous  equations,  181. 
Six-figure  accuracy,  18. 
Square  of  difference,  120. 

of  a  number,  57,  110. 

of  sum,  120. 
Square  rod  =  30j  square  yards. 

(160 square  rods  in  an  acre),  206. 
Square  root,  57,  110. 

in  equations,  65. 
Standard  parabola,  215. 
Straight-line  locus,  176. 
Straight  lines,  pair  of,  214. 
Straight  products,  122. 
Strip,  area  of,  64. 
Stripe,  32. 

figures  inscribed  in  a,  67. 
Study  of  a  simple  equation,  41. 
Subtraction,  rule  for,  107. 

of  fractions,  88. 

of  parentheses,  88. 
Subtrahends  in  division,  55. 
Suffixes,  32. 

Sum  of  the  angles  of  a  polygon, 
164. 

of    the    angles   of   a   triangle, 
33. 


Sum  of  the  base  angles  of  a  tri- 
angle, 33. 
of  the  unknown  numbers  given, 
44. 

Summation,  104. 

Supplementary  arcs,  47. 

Supplements,  23. 

Tangent  of  an  angle,  160. 

Tangents,  table  of,  160. 

Terms    in    algebraic    expressions, 

13. 
Theorem,  33,  119. 
Theorem  A,  120. 
Theorem  of  Pythagoras,  162. 
Three-figure  accuracy,  18. 

checking,  62. 
Tracing  paper,  use  of,  71,  186,  215. 
Train  dispatcher's  diagram,  169. 
Transversal,  32. 
Trapezoids,  75. 
Trial  product,  51. 
Triangles,  70. 

having  one  angle  the  same,  151. 

inscribed  in  a  stripe,  148 

similar  if  equiangular,  156. 
Trinomial,  111. 

Unit  of  length,  17. 
Unit-wide  strip,  area  of,  64. 

Valuation  problems,  26. 
Velocities,  problem  of  the  two,  94. 
Vertex  of  an  angle,  23. 
Vinculum,  108. 

Weight  of  air,  20,  60. 
of  carbon  dioxide,  60. 
of  coal,  53. 

Zero  obtained  by  adding,  104. 


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